# Semi-Supervised Deep Ensembles for Blind Image Quality Assessment

Zhihua Wang1111Contact Author Dingquan Li2 Kede Ma1
1Department of Computer Science, City University of Hong Kong
2 Peng Cheng Laboratory
zhihua.wang@my.cityu,edu.hk<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Ensemble methods are generally regarded to be better than a single model if
the base learners are deemed to be “accurate” and “diverse.” Here we
investigate a semi-supervised ensemble learning method to produce
generalizable blind image quality assessment models. We train a multi-head
convolutional network for quality prediction by maximizing the accuracy of the
ensemble (as well as the base learners) on labeled data, and the disagreement
(i.e., diversity) among them on unlabeled data, both implemented by the
fidelity loss. We conduct extensive experiments to demonstrate the advantages
of employing unlabeled data for BIQA, especially in model generalization and
failure identification.

##  1 Introduction

Data-driven blind image quality assessment (BIQA) models Bosse et al. (2018);
Ma et al. (2018) employing deep convolutional networks (ConvNets) have
achieved unprecedented performance, as measured by the correlations with human
perceptual scores. However, the performance improvement may be doubtful due to
the conflict between the small scale of test IQA datasets and the large scale
of ConvNet parameters, heightening the danger of poor generalizability.
Ensemble learning, which aims to improve model generalizability by making use
of multiple “accurate” and “diverse” base learners Zhang and Ma (2012), is a
promising way of alleviating this conflict, and has great potentials in
outputting generalizable BIQA models.

Researchers in the field of machine learning have contributed many brilliant
ideas to ensemble learning, e.g., bagging Breiman (1996), boosting Freund and
Schapire (1997), and negative correlation learning Liu and Yao (1999). These
methods are mainly demonstrated under the supervised learning setting where
the training labels are given. In many real-world problems, plenty of
unlabeled data are mostly freely accessible, while the collection of human
labels is prohibitively labor-expensive. For example, in BIQA, acquiring the
mean opinion score (MOS) of one image involves effort of $15$ to $30$ subjects
Sheikh et al. (2006).

In this paper, we combine labeled and unlabeled data to train deep ensembles
for BIQA in the semi-supervised setting Chapelle et al. (2006); Chen et al.
(2018). Our method is based on a multi-head ConvNet, where each head
corresponds to a base learner and produces a quality estimate. To reduce model
complexity, the base learners share considerable amount of early-stage
computation and have separate later-stage convolution and fully connected (FC)
layers Lee et al. (2015). The ensemble is end-to-end optimized to trade off
two theoretical conflicting objectives - ensemble accuracy (on labeled data)
and diversity (on unlabeled data), both implemented through the fidelity loss
Tsai et al. (2007). We conduct extensive experiments to show that the learned
ensemble performs favorably against a “top-performing” BIQA model - UNIQUE
Zhang et al. (2021b) in terms of quality prediction on existing IQA datasets,
while exhibiting much stronger generalizability in the group maximum
differentiation (gMAD) competition Ma et al. (2020). Moreover, our results
further suggest that promoting diversity helps the ensemble spot its corner-
case failures, which is in turn beneficial for subsequent active learning Wang
and Ma (2021).

##  2 Method

Let $f^{(i)}$ denotes the $i$-th base learner, which is implemented by a deep
ConvNet, consisting of several stages of convolution, batch normalization
Ioffe and Szegedy (2015), half-wave rectification (i.e., ReLU nonlinearity)
and spatial subsampling, followed by FC layers for quality computation. Given
an input image $x$, the ensemble method $f$ is simply defined by the average
of all base learners:

$\displaystyle f(x)=\frac{1}{M}\sum_{i=1}^{M}f^{(i)}(x),$ (1)

where $M$ is the number of base learners used to build the ensemble. A
necessary condition for Eq. (1) to be valid is that all base learners need to
produce quality estimates of the same perceptual scale, which is nontrivial to
satisfy when formulating BIQA as a ranking problem Ma et al. (2017). We will
give a detailed treatment of this scale alignment in Sec. 2.1.

###  2.1 Supervised Ensemble Learning for BIQA

Following Ma et al. (2017); Zhang et al. (2021b), we choose the pairwise
learning-to-rank (L2R) method for BIQA learning due to its feasibility of
training BIQA models on multiple human-rated datasets. Specifically, given a
labeled training set $\mathcal{L}$, in which each image $x$ is associated with
an MOS $\mu_{x}$, we convert it into another set $\mathcal{P}_{l}$, where the
input is a pair of images $(x,y)$ and the target output is a binary label

$\displaystyle p=\begin{cases}1&\mu_{x}\geq\mu_{y}\\\
0&\text{otherwise}\end{cases}.$ (2)

Under the Thurstone’s Case V model Thurstone (1927), we assume that the true
perceptual quality $q(x)$ of image $x$ follows a Gaussian distribution with
mean estimated by the ensemble $f(x)$. The probability $p(x,y)$ indicating
that $x$ is of higher quality than $y$ can then be computed by

$\displaystyle\hat{p}(x,y)=\operatorname{Pr}(q(x)\geq
q(y))=\Phi\left(\frac{f(x)-f(y)}{\sqrt{2}}\right),$ (3)

where $\Phi(\cdot)$ is the standard Normal cumulative distribution function,
and the standard deviation (std) is fixed to one. We may alternatively use the
$i$-th base learner to estimate the mean of the Gaussian distribution, and
compute a corresponding $\hat{p}^{(i)}(x,y)$ by replacing $f$ with $f^{(i)}$
in Eq. (3). We use the fidelity loss Tsai et al. (2007) to quantify the
similarity between two discrete probability distributions:

$\displaystyle\ell(p,\hat{p})=1-\sqrt{p\hat{p}}-\sqrt{(1-p)(1-\hat{p})}.$ (4)

We then define the optimization objective over a mini-batch
$\mathcal{B}_{l}\subset\mathcal{P}_{l}$ as

$\displaystyle\begin{split}\ell_{\mathrm{acc}}(\mathcal{B}_{l})=\frac{1}{\left|\mathcal{B}_{l}\right|}\sum_{x,y\in\mathcal{B}_{l}}&\Bigg{(}\ell\left(p(x,y),\hat{p}(x,y)\right)\Bigg{.}\\\
&+\Bigg{.}\frac{\lambda}{M}\sum_{i=1}^{M}\ell\left(p(x,y),\hat{p}^{(i)}(x,y)\right)\Bigg{)},\end{split}$
(5)

where the first term is the ensemble loss and the second term is the mean
individual loss, respectively. $\left|\mathcal{B}_{l}\right|$ denotes the
cardinality of $\mathcal{B}_{l}$. $\lambda$ is set to one by default.

It is noteworthy that the objective in Eq. (5) that relies on the Thurstone’s
model in Eq. (3) suffers from the translation ambiguity. As a result, the
learned base models $\\{f^{(i)}\\}_{i=1}^{M}$ may not live in the same
perceptual scale. We empirically find three simple tricks that are effective
in calibrating the base learners. First, $\ell_{2}$-normalize the input
feature vector to FC layers Wang et al. (2017); Zhang et al. (2021a),
projecting it onto the unit sphere. Second, remove the biases of the FC
layers. Third, batch-normalize the output of the FC layers (i.e., the quality
estimate) and share the learnable scale parameter across all base learners
(with the bias parameter fixed to zero).

As an additional note, an alternative way of computing $\hat{p}(x,y)$ by the
ensemble is to average the probabilities estimated by the base learners:

$\displaystyle\hat{p}(x,y)=\frac{1}{M}\sum_{i=1}^{M}\hat{p}^{(i)}(x,y).$

Throughout the paper, we opt for Eq. (3) because it is the ensemble prediction
(in Eq. (1)) that will be used during model deployment, and it gives slightly
better quality prediction results.

###  2.2 Semi-Supervised Ensemble Learning for BIQA

We now incorporate unlabeled data for learning BIQA models with the goal of
maximizing ensemble diversity. Similar to the supervised setting, we create a
second image set $\mathcal{P}_{u}$ by sampling pairs of images from a large
pool of unlabeled images. Inspired by the seminal work of negative correlation
learning Buschjäger et al. (2020); Chen et al. (2018), we define diversity as
the negative average of the pairwise fidelity losses between all pairs of base
learners:

$\displaystyle\ell_{\mathrm{div}}(\mathcal{B}_{u})=-\frac{1}{\binom{M}{2}\left|\mathcal{B}_{u}\right|}\sum_{x,y\in\mathcal{B}_{u}}\sum_{i<j}\ell\left(\hat{p}^{(i)}(x,y),\hat{p}^{(j)}(x,y)\right),$
(6)

where $\mathcal{B}_{u}$ is a mini-batch sampled from $\mathcal{P}_{u}$. Only
unordered pairs of base learners are considered due to the symmetry of the
fidelity loss (Eq. (4)). As there is no standardized definition of diversity,
other implementations may also be plausible, including the prediction
variance:

$\displaystyle\ell_{\mathrm{div}}(x)=-\frac{1}{M}\sum_{i=1}^{M}\left(f^{(i)}(x)-f(x)\right)^{2},$
(7)

where $f(x)$ is defined in Eq. (1) and the negative mean of the fidelity loss
between the base learners and the ensemble:

$\displaystyle\ell_{\mathrm{div}}(x,y)=-\frac{1}{M}\sum_{i=1}^{M}\ell\left(\hat{p}^{(i)}(x,y),\hat{p}(x,y)\right),$

where $\hat{p}(x,y)$ is defined in Eq. (3).

We combine the ensemble accuracy term on labeled data and the ensemble
diversity term on unlabeled data to obtain the final objective function:

$\displaystyle\ell_{\mathrm{semi}}(\mathcal{B}_{l},\mathcal{B}_{u})=\ell_{\mathrm{acc}}(\mathcal{B}_{l})+\gamma\ell_{\mathrm{div}}(\mathcal{B}_{u}),$
(8)

where $\gamma$ is the trade-off parameter. In our experiments, it does not
hurt to include a diversity term on the labeled $\mathcal{B}_{l}$ treated as
unlabeled Wang et al. (2021).

##  3 Experiments

In this section, we first describe the experimental setup, and then present
the main results, followed by extensive ablation studies.

###  3.1 Experimental Setups

Implementation Details. We use the convolutional structure in ResNet-18 He et
al. (2016) as the backbone, and add one FC layer for multi-head prediction.
The first convolution and the subsequent three residual blocks are shared for
each base learner to reduce the model complexity and computational cost Lee et
al. (2015). The backbone parameters are initialized with the weights pre-
trained on ImageNet Deng et al. (2009), and the FC parameters are initialized
by He’s method He et al. (2015). We train the entire method using the Adam
optimizer Kingma and Ba (2015) with a mini-batch size of $16$ for twelve
epochs. The initial learning rate is set to $10^{-4}$, which is halved for
every epoch. The best parameters are selected according to the performance on
the validation set. The images during training are cropped to $384\times 384$,
keeping their aspect ratios, while the test images are fed with the original
sizes.

(a)

(b)

(c)

(d)

Figure 1: Representative gMAD pairs between UNIQUE Zhang et al. (2021b) and the SSL ensemble on FLIVE Ying et al. (2020). Shown in the sub-caption is the MOS of each image. (a) Fixing UNIQUE at the low-quality level. (b) Fixing UNIQUE at the high-quality level. (c) Fixing the SSL ensemble at the low-quality level. (d) Fixing the SSL ensemble at the high-quality level.  Table 1: Correlation between model predictions and MOSs on the test sets of KonIQ-10k Hosu et al. (2020), SPAQ Fang et al. (2020), and FLIVE Ying et al. (2020), respectively. SRCC | KonIQ-10k | SPAQ | FLIVE
---|---|---|---
UNIQUE | $0.860$ | $0.899$ | $0.426$
Naïve Ensemble | $0.855$ | $0.900$ | $0.427$
Joint Ensemble | $0.864$ | $0.899$ | $0.430$
SSL Ensemble | $0.861$ | $0.900$ | $0.438$
PLCC | KonIQ-10k | SPAQ | FLIVE
UNIQUE | $0.859$ | $0.904$ | $0.505$
Naïve Ensemble | $0.859$ | $0.903$ | $0.517$
Joint Ensemble | $0.870$ | $0.903$ | $0.521$
SSL Ensemble | $0.861$ | $0.904$ | $0.527$
Table 2: Comparison of the failure-spotting efficiency as a function of the number of selected samples on FLIVE Ying et al. (2020). A lower correlation coefficient indicates better performance. SRCC | $500$ | $1,000$ | $1,500$ | $2,000$
---|---|---|---|---
UNIQUE | $0.358$ | $0.373$ | $0.352$ | $0.377$
Naïve Ensemble | $0.458$ | $0.397$ | $0.391$ | $0.384$
Joint Ensemble | $0.346$ | $0.355$ | $0.360$ | $0.370$
SSL Ensemble | $0.338$ | $0.305$ | $0.318$ | $0.326$
PLCC | $500$ | $1,000$ | $1,500$ | $2,000$
UNIQUE | $0.463$ | $0.460$ | $0.456$ | $0.483$
Naïve Ensemble | $0.504$ | $0.461$ | $0.452$ | $0.441$
Joint Ensemble | $0.483$ | $0.483$ | $0.488$ | $0.482$
SSL Ensemble | $0.378$ | $0.384$ | $0.390$ | $0.400$

Datasets. We use $60\%$ images in KonIQ-10k Hosu et al. (2020) and SPAQ Fang
et al. (2020) as the labeled training set, $20\%$ as the validation set, and
$20\%$ as the test set, respectively. We treat the entire FLIVE Ying et al.
(2020) as the unlabeled training set. To reduce the bias caused by the random
splitting, we repeat the procedure three times, and report the mean results.

Evaluation Metrics. We adopt two quantitative criteria: Spearman rank-order
correlation coefficient (SRCC) and Pearson linear correlation coefficient
(PLCC), to measure prediction performance, respectively. Before computing
PLCC, a four-parameter logistic function is suggested in VQEG (2000) to fit
model predictions to subjective scores:

$\displaystyle\hat{f}(x)=(\eta_{1}-\eta_{2})/(1+\exp(-(f(x)-\eta_{3})/|\eta_{4}|))+\eta_{2},$
(9)

where $\\{\eta_{i};i=1,2,3,4\\}$ are the parameters to be optimized.

###  3.2 Main Results

Correlation Results. We compare our method (termed as the SSL ensemble)
against three variants: 1) UNIQUE Zhang et al. (2021b), a state-of-the-art
BIQA model, corresponding to $M=1$; 2) Naïve Ensemble, corresponding to
$\lambda=0$ in Eq. (5) and $\gamma=0$ in Eq. (8); 3) Joint Ensemble,
corresponding to $\gamma=0$. The mean SRCC and PLCC results are listed in
Table 1, where we find that the SSL ensemble performs favorably against the
competing methods.

Table 3: The effect of the number of base learners in the SSL ensemble. The default setting is highlighted in bold. SRCC | KonIQ-10k | SPAQ | FLIVE
---|---|---|---
$2$ | $0.859$ | $0.897$ | $0.418$
$4$ | $0.861$ | $0.896$ | $0.421$
$6$ | $0.861$ | $0.897$ | $0.424$
8 | $0.861$ | $0.900$ | $0.438$
$12$ | $0.864$ | $0.900$ | $0.426$
PLCC | KonIQ-10k | SPAQ | FLIVE
$2$ | $0.854$ | $0.901$ | $0.506$
$4$ | $0.862$ | $0.900$ | $0.509$
$6$ | $0.863$ | $0.900$ | $0.513$
8 | $0.861$ | $0.904$ | $0.527$
$12$ | $0.871$ | $0.903$ | $0.518$

gMAD Results. We next let the SSL ensemble play the gMAD competition game Ma
et al. (2020) with UNIQUE on the FLIVE database. Figure 1 shows the
representative gMAD pairs. It is clear that the pairs of images in (a) and
(b), where UNIQUE and the SSL ensemble perform the defender and attacker
roles, respectively, exhibit substantially different quality, which is in
disagreement with UNIQUE. In contrast, the SSL ensemble correctly predicts the
top images to have much better quality than the corresponding bottom images.
When switching their roles (see (c) and (d)), we find that the SSL ensemble
successfully survives the attacks from UNIQUE, with pairs of images of similar
quality according to human perception. This provides a strong indication that
the SSL ensemble leads to better generalization to novel images.

Failure Identification Results. It is of great importance to efficiently spot
the catastrophic failures of “top-performing” BIQA models, offering the
opportunity to further improve the models. Wang and Ma Wang and Ma (2021) made
one of the first attempts to identify the counterexamples of UNIQUE Zhang et
al. (2021b) with the help of multiple full-reference IQA models. A significant
advantage of the proposed ensemble is that it comes with a natural failure
identification mechanism: we may seek samples by the principle of maximal
disagreement, as measured by Eq. (7). This is known as the query by committee
Seung et al. (1992) in the active learning literature. To enable comparison
with UNIQUE Zhang et al. (2021b), we re-train it with the dropout technique
Srivastava et al. (2014) and use Monte Carlo dropout Gal and Ghahramani (2016)
to approximate Eq. (7) for failure identification. Table 2 lists the SRCC and
PLCC results as a function of the sample size on FLIVE. Here lower correlation
indicates higher failure-spotting efficiency. As can be seen, the SSL ensemble
achieves the lowest correlation across all sample sizes, providing strong
justifications of exploiting unlabeled data for diversity promotion. We are
currently testing the use of the spotted samples for model refinement in a
similar active fine-tuning framework described in Wang and Ma (2021).
Preliminary results show that the samples with the maximal disagreement are
beneficial for improving the model generalizability.

Table 4: The effect of the trade-off parameter ($\gamma$ in Eq. (8)) for training the SSL ensemble. SRCC | KonIQ-10k | SPAQ | FLIVE
---|---|---|---
$0.00$ | $0.855$ | $0.900$ | $0.427$
$0.04$ | $0.864$ | $0.899$ | $0.429$
0.06 | $0.861$ | $0.900$ | $0.438$
$0.08$ | $0.860$ | $0.895$ | $0.432$
$0.10$ | $0.851$ | $0.894$ | $0.437$
PLCC | KonIQ-10k | SPAQ | FLIVE
$0.00$ | $0.870$ | $0.903$ | $0.521$
$0.04$ | $0.868$ | $0.903$ | $0.518$
0.06 | $0.861$ | $0.904$ | $0.527$
$0.08$ | $0.865$ | $0.899$ | $0.522$
$0.10$ | $0.848$ | $0.897$ | $0.525$
Table 5: The effect of different splitting points in the SSL ensemble using ResNet-18 as the backbone. The configuration in the first column indicates the layer, after which the computation is not shared. SRCC | KonIQ-10k | SPAQ | FLIVE
---|---|---|---
First Convolution | $0.866$ | $0.899$ | $0.434$
Residual Block $1$ | $0.865$ | $0.899$ | $0.434$
Residual Block $2$ | $0.864$ | $0.899$ | $0.432$
Residual Block 3 | $0.861$ | $0.900$ | $0.438$
Residual Block $4$ | $0.864$ | $0.900$ | $0.437$
PLCC | KonIQ-10k | SPAQ | FLIVE
First Convolution | $0.871$ | $0.903$ | $0.525$
Residual Block $1$ | $0.869$ | $0.903$ | $0.525$
Residual Block $2$ | $0.869$ | $0.903$ | $0.523$
Residual Block 3 | $0.861$ | $0.904$ | $0.527$
Residual Block $4$ | $0.864$ | $0.904$ | $0.525$

###  3.3 Ablation Studies

In this subsection, we perform ablation experiments to examine the influence
of three hyper-parameters: 1) the number of base learners, $M$, 2) the trade-
off parameter for ensemble accuracy and diversity, $\bm{\gamma}$, and 3) the
splitting point, before which all base learners share the computation.

Table 3 shows the effect of employing different numbers of base learners
(i.e., $M\in\\{2,4,6,8,12\\}$) in the SSL ensemble, where we find that our
method is fairly stable with regard to this hyperparameter, and adding more
base learners slightly increase the correlation numbers. We next test the
influence of the trade-off parameter $\gamma$ in Eq. (8). Table 4 shows the
performance change with $\gamma$ sampled from
$\\{0.00,0.04,0.06,0.08,0.10\\}$. It is not surprising that a larger value
(e.g., $\gamma=0.10$) may harm the ensemble performance, especially on
KonIQ-10k. This is because excessive diversity may destroy the prediction
accuracy of base learners and (subsequently) the ensemble Brown and Kuncheva
(2010). This phenomenon would be more pronounced if the adopted diversity
measure is unbounded from above (e.g., Eq. (7)). In general, the correlation
is not sensitive when $\gamma$ is relatively small. Last, we explore the
influence of the splitting point determining how much computation is shared by
the base learners. From the Table 5, we observe stable performance at
different splitting points. Therefore, it is encouraged to share more layers
to reduce the inference time and the memory requirement.

##  4 Conclusions

In this paper, we have conducted an empirical study to probe the advantages of
incorporating unlabeled data for training BIQA ensembles. This naturally leads
to a semi-supervised formulation, where we maximized the ensemble accuracy on
labeled data and the ensemble diversity on unlabeled data. Through
comprehensive experiments, we arrived at three interesting findings. First,
the diversity-driven SSL ensemble does not achieve correlation improvements on
existing IQA databases. Second, despite similar correlation performance, our
ensemble shows much stronger generalizability in the gMAD competition with
greater potentials for use in monitoring real-world quality problems. Third,
our ensemble has a built-in failure-identification mechanism with demonstrated
efficiency. This points to an interesting avenue for future work - active
fine-tuning BIQA models in the semi-supervised setting.

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#  New applications to combinatorics and invariant matrix norms of an integral
representation of natural powers of the numerical values

Hassan Issa∗†‡ ,  Bassam Mourad∗ and  Hassan Abbas∗ *Department of
Mathematics, Faculty of Science I, Lebanese University, Beirut, Lebanon.  †
first author: Hassan A. Issa<EMAIL_ADDRESS>
(Date: May 23, 2021)

###### Abstract.

Let $\vee^{k}A$ be the $k$-th symmetric tensor power of $A\in
M_{n}(\mathbb{C})$. In [23], we have expressed the normalized trace of
$\vee^{k}A$ as an integral of the $k$-th powers of the numerical values of $A$
over the unit sphere $\mathbb{S}^{n}$ of $\mathbb{C}^{n}$ with respect to the
normalized Euclidean surface measure $\sigma$. In this paper, we first use
this integral representation to construct a family of unitarily invariant
norms on $M_{n}(\mathbb{C})$ and then explore their relations to Schatten-
norms of $\vee^{k}A$. Another application yields a connection between the
analysis of symmetric gauge functions with that of complete symmetric
polynomials. Finally, motivated by the work of R. Bhatia and J. Holbrook in
[10], and as pointed out by R. Bhatia in [6] in the development of the theory
of weakly unitarily invariant norms, we provide an explicit form for the
weakly unitarily invariant norm corresponding to the $L^{4}$-norm on the space
$C(\mathbb{S}^{n})$ of continuous functions on the sphere. Our result
generalize those of R. Bhatia and J. Holbrook in different directions and pave
the way to a technique for computing those weakly unitarily invariant norms on
$M_{n}(\mathbb{C})$ that are associated to $L^{2k}$-norms on
$C(\mathbb{S}^{n})$.

###### Key words and phrases:

Integral formula, trace of symmetric product, unitarily invariant norms,
weakly unitarily invariant norms.

###### 2010 Mathematics Subject Classification:

Primary 15A15, 15A60; Secondary 32A26

##  1\. Introduction

For $A\in M_{n}(\mathbb{C})$, let $trA$ be its trace and $\vee^{k}A$ denote
its $k$-th symmetric tensor product where $k\in\mathbb{N}$. In addition, let
$Tr(\vee^{k}A)=\dfrac{tr(\vee^{k}A)}{c_{n,k}}$ be the normalized trace of
$\vee^{k}A$ with $c_{n,k}=\binom{n+k-1}{k}$. In [23], we have expressed
$Tr(\vee^{k}A)$ as an integral of the $k$-th powers of the numerical values of
$A$ over the unit sphere $\mathbb{S}^{n}$ of $\mathbb{C}^{n}$ with respect to
the normalized Euclidean surface measure $\sigma$. More precisely, we have
proved the following.

$\int_{\mathbb{S}^{n}}\big{(}<A\xi,\xi>\big{)}^{k}d\sigma(\xi)=Tr(\vee^{k}(A))=\frac{1}{c_{n,k}}\sum_{l=1}^{k}\sum_{\beta\in
S^{k}_{l}}\frac{1}{z_{\beta}}\prod_{t=1}^{l}tr(A^{\beta_{t}}),$ (1.1)

where

$S^{k}_{l}=\\{\beta=(\beta_{1},\beta_{2},\cdots,\beta_{k})\in\mathbb{N}^{k}\mid\beta_{1}\geq\beta_{2}\geq\cdots\geq\beta_{k},|\beta|=k,\beta_{l}\neq
0\mbox{ and }\beta_{i}=0\mbox{ for all }i>l\\}.$

Since $Tr(\vee^{k}A)$ is simply the evaluation of the normalized complete
$k$-homogenous polynomials $H_{k}$ at the eigenvalues of $A$, the preceding
integral can be then interpreted as a representation for such polynomials.

In this paper, we continue to explore the potential consequences of (1.1) in
many directions. For convenience, we shall use the same terminology and
notation as in [23] and we shall always assume that $k$ is non-zero.

Recall that the Hilbert-Schmidt norm $\lVert\cdot\rVert_{(2)}$ of the integral
transform $T_{A,k}$ on $L^{2}(\mathbb{S}^{n})$ satisfies

$c^{\frac{1}{2k}}_{n,k}\lVert
T_{A,k}\rVert^{\frac{1}{k}}_{(2)}=\left(\int_{\mathbb{S}^{n}}\lVert
A\xi\rVert_{2}^{2k}d\sigma(\xi)\right)^{\frac{1}{2k}}.$ (1.2)

From the measure theory point of view, this can be regarded as the
$L^{2k}(\mathbb{S}^{n},\mathbb{C}^{n})$ Bochner norm of the map
$\mathbb{S}^{n}\ni\xi\longrightarrow A\xi\in\mathbb{C}^{n}$, whenever
$\mathbb{C}^{n}$ is equipped with $\lVert\cdot\rVert_{2}$. A slightly careful
argument leads to a norm on $M_{n}(\mathbb{C})$ but with some extremal
property of unitary invariance. Note that, a norm $\tau$ on
$M_{n}(\mathbb{C})$ is called unitarily invariant (u.i. for short) if it
satisfies

$\tau(A)=\tau(UAV),\quad\mbox{for all}\quad A\in
M_{n}(\mathbb{C})\quad\mbox{and all}\quad U,V\in U(n),$ (1.3)

where $U(n)$ is the group of all unitary matrices in $M_{n}(\mathbb{C})$.

A detailed description of this paper is presented as follows. In Section 2, we
shall take advantage of Thompson triangular inequality to prove that for each
$k\in\mathbb{N}$ the right hand side of (1.2) defines a Q-norm i.e. a norm for
which there exists a unitarily invariant norm $N_{k}$ on $M_{n}(\mathbb{C})$
such $\sqrt{N_{k}(A^{\star}A)}$ is given by (1.2) (cf. [7, 8]). Using (1.1),
we provide an expression for $N_{k}$ as a non-trivial combination of Schatten
p-norms $\lVert\cdot\rVert_{(p)}$ on $M_{n}(\mathbb{C})$. A connection to a
Schatten norm of the symmetric power is also explored.

Characterized by Von-Neumann in [35], unitarily invariant norms are special
norms on the sequence of singular values and this characterization is
clarified via the concept of symmetric gauge functions. Recall that a
symmetric gauge function (s.g.f..) is a norm on $\mathbb{R}^{n}$ being
invariant under the permutation and the change of signs of the coordinates. Of
special interest are the inequalities satisfied by these functions (cf. [6]
Ch. IV and the references therein). In Section 3, we apply such inequalities
to explicitly construct those unitarily invariant norms studied in Section 2.
This yields various simple proofs for the analysis of the complete symmetric
polynomials $h_{k}$, and this approach is apparently novel in combinatorics
and has the advantage of giving a new look at the topic from a different
perspective.

The last section is devoted to the investigation of a problem on weakly
unitarily invariant (w.u.i.) norms which dates back to the work of R. Bhatia
and J. Holbrook in [10] in 1987. Recall that a norm $\tau$ on
$M_{n}(\mathbb{C})$ is called weakly unitarily invariant norm if it satisfies
the invariant property

$\tau(A)=\tau(UAU^{\star}),\quad\mbox{for all}\quad A\in
M_{n}(\mathbb{C})\quad\mbox{and all}\quad U\in U(n),$ (1.4)

where $U^{\star}$ denotes as usual the conjugate transpose of $U$. Such norms
that have been studied by C.-K.Fong and J.A.R. Holbrook in [16] on arbitrary
dimensional Hilbert spaces, were characterized by Chi-K. Li and N. K. Tsing in
[26] as Schur-convex norms (rather than symmetric gauge functions for u.i.
norms) when considered as maps on the cone of Hermitian matrices
$\mathcal{H}_{n}$. However, R. Bhatia and J.A.R. Holbrook ([10]) provided a
complete characterization for w.u.i. norms in terms of ”unitary function
norms” on the space $C(\mathbb{S}^{n})$ of continuous functions on the sphere.
Following [10], a norm $\Phi$ on $C(\mathbb{S}^{n})$ is called unitarily
invariant if

$\Phi(f)=\tau(f\circ U),\quad\mbox{for all}\quad f\in
C(\mathbb{S}^{n})\quad\mbox{and all}\quad U\in U(n).$ (1.5)

Given a norm $\Phi$ on $C(\mathbb{S}^{n})$, let
$\Phi^{\prime}:M_{n}(\mathbb{C})\longrightarrow\mathbb{R}$ be the map defined
by

$\Phi^{\prime}(A)=\Phi(f_{A}),\quad\mbox{ where
}f_{A}:\mathbb{S}^{n}\longrightarrow\mathbb{C}\quad\mbox{is given by:
}f_{A}(\xi)=<A\xi,\xi>.$ (1.6)

More precisely, their characterization is given as follows: A norm $\tau$ on
$M_{n}(\mathbb{C})$ is w.u.i if and only if there is a unitarily invariant
norm $\Phi$ on $C(\mathbb{S}^{n})$ such that $\tau=\Phi^{\prime}$ (cf. [10],
Theorem 2.1).

This explicit correspondence between $\tau$ and $\Phi$ could lead to a more
development of the theory of w.u.i. norms and such correspondence is ”not
known for even much used norms” ([6], p. 110). In particular, this holds for
the natural $L^{p}$-norms $\Phi_{p}$ on $C(\mathbb{S}^{n})$ (w.r.t $\sigma$)
with $p\in[1,\infty[\backslash\\{2\\}$. It is direct to see that
$\Phi^{\prime}_{\infty}$ is the numerical radius norm. The case when $p=2$,
was obtained by R. Bhatia and J. Holbrook (cf. [10], Theorem 2.4) and
explicitly it reads as

$\Phi^{\prime}_{2}(A)=\Bigg{(}\frac{\lVert
A\rVert_{F}^{2}+|tr(A)|^{2}}{c_{n,2}}\Bigg{)}^{\frac{1}{2}}.$ (1.7)

We shall make use of (1.1) to present the exact values for the following:

  1. (1)

$\Phi^{\prime}_{k}$ with $k\in\mathbb{N}$, on the cone of positive semi-
definite matrices $\mathcal{H}_{n}^{+}$,

  2. (2)

$\Phi^{\prime}_{2k}$ on $\mathcal{H}_{n}$,

  3. (3)

$\Phi^{\prime}_{4}$ on $M_{n}(\mathbb{C})$.

Furthermore, we shall modify $\Phi^{\prime}_{4}$ in order to obtain a new
w.u.i. norm with some probabilistic interpretation. Our result generalizes the
inner product form on $M_{n}(\mathbb{C})$ obtained in [10]. Finally. we
conclude with a method that produces the exact value for $\Phi^{\prime}_{2k}$
on $M_{n}(\mathbb{C})$.

##  2\. On a class of unitarily invariant norms

In this section, we shall first provide a collection of unitarily invariant
norms on $M_{n}(\mathbb{C})$ indexed by real numbers $q\geq 1$. In the case
where $q$ is an integer, we shall show that these norms are formed of a non-
trivial combination of products of Schatten $p$-norms. Then, we shall focus on
a family of Q-norms of the form (1.2) which will be used later in our
investigation for the weakly unitarily invariant norms in the last section.

In the sequel, for $A\in M_{n}(\mathbb{C})$ and $p\in[1,\infty[$ we shall
write $|A|:=\sqrt{A^{\star}A}$ for the positive part of $A$,
$\lambda:=\lambda(A)$ for the unordered tuple of eigenvalues of $A$,
$\sigma(A):=\lambda(|A|)\in\mathbb{R}^{n}_{+}$ and $\lVert
A\rVert_{(p)}:=\big{[}tr(|A|^{p})\big{]}^{\frac{1}{p}}$ for the Schatten
$p$-norms. Moreover, to indicate that $A\in\mathcal{H}_{n}^{+}$ we simply
write $A\geq 0$ and as usual for $A,B\geq 0$ we write $A\geq B$ if $A-B\geq
0.$

######  Definition 2.1.

Given $q\in[1,\infty[$ and $n\in\mathbb{N}$, we define the map
$N_{q}:M_{n}(\mathbb{C})\longrightarrow\mathbb{R}_{+}$ by

$N_{q}(A):=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}},$
(2.1)

where $\lVert\cdot\rVert_{2}$ denotes the Euclidean norm on $\mathbb{C}^{n}.$

We aim to show that each $N_{q}$ defines a unitarily invariant norm on
$M_{n}(\mathbb{C})$. Even though, Formula (2.1) can be viewed as the
$L^{q}(\mathbb{S}^{n})$-norm for the map
$\xi\longrightarrow\lVert|A|^{\frac{1}{2}}\xi\rVert_{2}^{2}$, this is not
sufficient to obtain Minkowski’s inequality. When $q=k$ is an integer, then by
(1.1) we can write

$N_{k}(A)=\Big{[}Tr\big{(}\vee^{k}|A|\big{)}\Big{]}^{\frac{1}{k}},\quad A\in
M_{n}(\mathbb{C}).$ (2.2)

Now the convexity of the map
$A\longrightarrow\big{[}Tr(\vee^{k}(A))\big{]}^{\frac{1}{k}}$ on the set of
positive matrices $\mathcal{H}_{n}$ certainly ensures that $N_{k}(A+B)\leq
N_{k}(A)+N_{k}(B)$ whenever $A,B\in\mathcal{H}^{+}_{n}$ (cf. [29, 31] and Ch.
II in [6]). Such approach is thus limited to $\mathcal{H}^{+}_{n}$. It is
worthy to mention here that one may combine the triangular inequality of Ky
Fan norms, the monotonicity and the convexity of the $k$th root of the
complete symmetric polynomials to solve such a problem for the integer case.
However, we will prove Minkowski’s inequality for $N_{q}$ for any $q\geq 1$
using Thompson triangular inequality together with the triangular inequality
of the $L^{q}(\mathbb{S}^{n})$-norm but without use of any other convexity
results in this discipline. Let us recall Thompson triangular inequality for
the positive parts of matrices.

######  Lemma 2.1 ( [40], Theorem 2 ).

Let $A,B\in M_{n}(\mathbb{C})$ then there are $U,V\in U(n)$ satisfying

$|A+B|\leq U|A|U^{\star}+V|B|V^{\star}.$ (2.3)

With the above notation, we now present our first result of this section.

######  Theorem 2.1.

For each $q\geq 1$ and each $n\in\mathbb{N}$, the map $N_{q}$ given by (2.1)
defines a unitarily invariant norm on $M_{n}(\mathbb{C})$.

###### Proof.

Let us first notice that for any $C\in M_{n}(\mathbb{C})$ and every $U,V\in
U(n)$, the following identity holds

$\displaystyle\int_{\mathbb{S}^{n}}\Big{\lVert}UCV\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)=\int_{\mathbb{S}^{n}}\Big{\lVert}C\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi).$
(2.4)

Indeed,

$\begin{split}\int_{\mathbb{S}^{n}}\Big{\lVert}UCV\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)&=\int_{\mathbb{S}^{n}}\Big{\lVert}CV\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\\\
&=\int_{\mathbb{S}^{n}}\big{(}<CV\xi,CV\xi>\big{)}^{q}d\sigma(\xi)=\int_{\mathbb{S}^{n}}\big{(}<C\mu,C\mu>\big{)}^{q}d\sigma(\mu),\end{split}$

where the first equality follows from the basic fact that unitary matrices
preserves the Euclidean norm and the last equality holds by the invariance of
$\sigma$ under a unitary transformation. The first two conditions for $N_{q}$
to be a norm on $M_{n}(\mathbb{C})$ are easy to obtain. For the triangular
inequality, we fix $A,B\in M_{n}(\mathbb{C})$ then by Lemma 2.1 there are
unitary matrices $U$ and $V$ satisfying (2.3). So that for any
$\xi\in\mathbb{S}^{n}$ we have

$\displaystyle\Big{\lVert}|A+B|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2}$
$\displaystyle=<|A+B|\xi,\xi>$
$\displaystyle\leq<U|A|U^{\star}\xi,\xi>+<V|B|V^{\star}\xi,\xi>$
$\displaystyle=\Big{\lVert}|A|^{\frac{1}{2}}U^{\star}\xi\Big{\rVert}_{2}^{2}+\Big{\lVert}|B|^{\frac{1}{2}}V^{\star}\xi\Big{\rVert}_{2}^{2}$

Considering the $L^{q}(\mathbb{S}^{n})$-norm for the functions on both sides
of the preceding inequality yields

$\displaystyle\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A+B|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}$
$\displaystyle\leq\left(\int_{\mathbb{S}^{n}}\Bigg{(}\Big{\lVert}|A|^{\frac{1}{2}}U^{\star}\xi\Big{\rVert}_{2}^{2}+\Big{\lVert}|B|^{\frac{1}{2}}V^{\star}\xi\Big{\rVert}_{2}^{2}\Bigg{)}^{q}d\sigma(\xi)\right)^{\frac{1}{q}}$
$\displaystyle\leq\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{1}{2}}U^{\star}\xi\Big{\rVert}_{2}^{2q}\Bigg{)}^{\frac{1}{q}}+\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|B|^{\frac{1}{2}}V^{\star}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}$
(2.5)
$\displaystyle=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}+\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|B|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}},$
(2.6)

where (2.5) follows from the triangular inequality of the
$L^{q}(\mathbb{S}^{n})$-norm and (2.6) follows by suitable applications of
(2.4). Next, to check the unitary invariance property (1.3) of $N_{q}$,
consider $A\in M_{n}(\mathbb{C})$ and $U,V\in U(n)$ then

$\displaystyle N_{q}(UAV)$
$\displaystyle=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|UAV|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|AV|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}$
(2.7)
$\displaystyle=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}V^{\star}|A|^{\frac{1}{2}}V\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}},$
(2.8)

where (2.7) follows from the fact that $|UC|=|C|$ and the first equality of
(2.8) is obtained by making use of the fact that $|CV|=V^{\star}|C|V$ for any
$C\in M_{n}(\mathbb{C})$. While the second equality of (2.8) follows easily
from (2.4). ∎

For the case $q=k\in\mathbb{N}$, we replace $A$ by $|A|$ in the second
equality of (1.1) and we obtain an expression for $N_{k}$ in terms of Schatten
norms as follows.

######  Corollary 2.1.

Let $A\in M_{n}(\mathbb{C})$ and let $k\in\mathbb{N}$ then the following holds

$c_{n,k}N_{k}^{k}(A)=\sum_{l=1}^{k}\sum_{\beta\in
S^{k}_{l}}\frac{1}{z_{\beta}}\prod_{t=1}^{l}\big{\lVert}A\big{\rVert}^{\beta_{t}}_{{}_{(\beta_{t})}}.$
(2.9)

Motivated by the above result and similar to Example 2.1 in [23], we express
$N_{k}$ in terms of Schatten norms for the cases $k=3,4$ and $5$ (the cases
$k=1$ or $k=2$ can be obtained directly from the results in [10]).

######  Example 2.1.

For a given $k,n\in\mathbb{N}$ and for any $A\in M_{n}(\mathbb{C})$ the
following identities hold []

$\begin{split}\mbox{i)}\quad c_{n,3}N_{3}^{3}(A)=&\frac{1}{3}\lVert
A\rVert^{3}_{(3)}+\frac{1}{2}\lVert A\rVert^{2}_{(2)}.\lVert
A\rVert_{(1)}+\frac{1}{6}\lVert A\rVert^{3}_{(1)}.\\\ \mbox{ii)}\quad
c_{n,4}N_{4}^{4}(A)=&\frac{1}{4}\lVert A\rVert^{4}_{(4)}+\frac{1}{3}\lVert
A\rVert^{3}_{(3)}\lVert A\rVert_{(1)}+\frac{1}{8}\lVert
A\rVert^{4}_{(4)}+\frac{1}{4}\lVert A\rVert^{2}_{(2)}\lVert
A\rVert^{2}_{(1)}+\frac{1}{24}\lVert A\rVert^{4}_{(1)}.\\\ \mbox{iii)}\quad
c_{n,5}N_{5}^{5}(A)=&\frac{1}{5}\lVert A\rVert^{5}_{(5)}+\frac{1}{4}\lVert
A\rVert^{4}_{(4)}\lVert A\rVert_{(1)}+\frac{1}{6}\lVert
A\rVert^{3}_{(3)}\lVert A\rVert^{2}_{(2)}+\frac{1}{6}\lVert
A\rVert^{3}_{(3)}\lVert A\rVert^{2}_{(1)}+\frac{1}{8}\lVert
A\rVert^{4}_{(2)}\lVert A\rVert_{(1)}\\\ &+\frac{1}{12}\lVert
A\rVert^{2}_{(2)}\lVert A\rVert^{3}_{(1)}+\frac{1}{120}\lVert
A\rVert^{5}_{(1)}.\end{split}$

Making use of Minkowski’s inequality for gauge functions (Theorem IV.1.8
in[6]), one can construct new unitarily invariant norms from a given u.i.
norm. In particular, we may apply such a construction on each $N_{q}$ to
obtain a more general u.i. norm. Let $q\geq 1$ be fixed, for each
$p\in[1,\infty[$ consider the map
$N_{q}^{(p)}:M_{n}(\mathbb{C})\longrightarrow\mathbb{R}_{+}$ defined by

$N_{q}^{(p)}(A):=N^{\frac{1}{p}}_{q}(|A|^{p})=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{p}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{pq}}=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{(}<|A|^{p}\xi,\xi>\Big{)}^{q}d\sigma(\xi)\Bigg{)}^{\frac{1}{pq}}.$
(2.10)

Then $N_{q}^{(p)}$ defines a u.i. norm on $M_{n}(\mathbb{C})$ (cf. for example
Ex. IV.2.7 in [6]). On the one hand, using a similar argument as the one used
in the previous discussion, we can conclude that if $q=k$ is an integer then
$N_{k}^{(p)}$ satisfies the following identity

$c_{n,k}\big{[}N_{k}^{(p)}(A)\big{]}^{pk}=\sum_{l=1}^{k}\sum_{\beta\in
S^{k}_{l}}\frac{1}{z_{\beta}}\prod_{t=1}^{l}\big{\lVert}A\big{\rVert}^{p\beta_{t}}_{{}_{(p\beta_{t})}},\quad
p\in[1,\infty[,A\in M_{n}(\mathbb{C}).$ (2.11)

On the other hand, by formulating the previous equation using the first
equality in (1.1), one obtains the ”surprising result” that the Schatten
p-norms (with a modified power) on the space of the $k$th-symmetric powers of
$M_{n}(\mathbb{C})$ provides a u.i. norm on $M_{n}(\mathbb{C})$. The following
proposition demonstrates this fact.

######  Proposition 2.1.

Let $k\in\mathbb{N}$ and $p\in[1,\infty[$, then

$c^{\frac{1}{kp}}_{n,k}N_{k}^{(p)}(A)=\Big{\lVert}\vee^{k}A\Big{\rVert}_{(p)}^{\frac{1}{k}},\quad
A\in M_{n}(\mathbb{C}).$ (2.12)

###### Proof.

From (1.1) we know that

$c_{n,k}\big{[}N_{k}^{(p)}(A)\big{]}^{pk}=c_{n,k}\int_{\mathbb{S}^{n}}\Big{(}<|A|^{p}\xi,\xi>\Big{)}^{k}d\sigma(\xi)=tr\Big{(}\vee^{k}|A|^{p}\Big{)}.$

Thus, it suffices to prove that

$tr\Big{(}\vee^{k}|A|^{p}\Big{)}=tr\Big{(}\big{|}\vee^{k}A\big{|}^{p}\Big{)}.$
(2.13)

Let us show first that for any $B\geq 0$ and any $\alpha\in\mathbb{R}_{+}$ the
following holds

$tr\Big{(}\vee^{k}(B^{\alpha})\Big{)}=tr\Big{(}(\vee^{k}B)^{\alpha}\Big{)}.$
(2.14)

The equality $\vee^{k}(B^{q})=(\vee^{k}B)^{q}$ for $q\in\mathbb{Q}_{+}$
follows from the fact that $\vee^{k}B\vee^{k}B=\vee^{k}B^{2}$ and by a simple
induction. Now for $\alpha\in\mathbb{R}_{+}$, choose a sequence
$(q_{m})\subset\mathbb{Q}_{+}$ converging to $\alpha$. By the continuity of
$\lambda$ as a map from $M_{n}(\mathbb{C})$ onto $\mathbb{C}^{n}$, we conclude
that the sequence of vectors

$\lambda(B^{q_{m}})=\big{(}\lambda_{1}(B^{q_{m}}),\cdots,\lambda_{n}(B^{q_{m}})\big{)}=\Big{(}\big{(}\lambda_{1}(B)\big{)}^{q_{m}},\cdots,\big{(}\lambda_{n}(B)\big{)}^{q_{m}}\Big{)}$

converges to

$\lambda(B^{\alpha})=\Big{(}\big{(}\lambda_{1}(B)\big{)}^{\alpha},\cdots,\big{(}\lambda_{n}(B)\big{)}^{\alpha}\big{)}.$

Now if $f_{k}$ denotes the $k$th complete symmetric polynomial on
$\mathbb{C}^{n}$, which is trivially continuous, then

$\displaystyle
tr\Big{(}\vee^{k}(B^{\alpha})\Big{)}=f_{k}(\lambda(B^{\alpha}))$
$\displaystyle=\lim_{m\rightarrow\infty}f_{k}\lambda(B^{q_{m}})=\lim_{m\rightarrow\infty}tr\big{(}\vee^{k}(B^{q_{m}})\big{)}$
$\displaystyle=\lim_{m\rightarrow\infty}tr\Big{(}(\vee^{k}B)^{q_{m}}\Big{)}=tr\Big{(}(\vee^{k}B)^{\alpha}\Big{)},$

where the last equality follows from the continuity of the trace on
$M_{c_{n,k}}(\mathbb{C})$. Now using the basic fact that
$\vee^{k}|A|=|\vee^{k}A|$, Eq. (2.13) then follows by applying (2.14) to the
matrix $B=|A|$. ∎

For the case $p=2$, we obtain a family of Q-norms $\\{N_{q}^{(2)}\\}_{q\geq
1}$ (see Def. IV 2.9 in [6]). Those norms will be used in Section 3 for the
analysis of the complete homogeneous and in Section 4 to obtain some upper
estimations for a family of w.u.i. norms. For simplicity, we shall write
$N^{\prime}_{q}$ rather than $N_{k}^{(2)}$ i.e.

$N^{\prime}_{q}(A)=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{2q}}=\Bigg{(}\int_{\mathbb{S}^{n}}\lVert
A\xi\rVert_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{2q}}.$ (2.15)

We conclude this section by presenting a proof for (1.2). Following the
notation in Section 1, we shall make use of the following well known formula
(cf. Theorem 3.5 in [41]) to compute the Hilbert-Schmidt norm of $T_{A,k}$:

$\big{\lVert}T_{A,k}\big{\rVert}^{2}_{(2)}=\int_{\mathbb{S}^{n}}\int_{\mathbb{S}^{n}}\big{|}<A\xi,\mu>\big{|}^{2k}d\sigma(\xi)d\sigma(\mu).$

With this in mind, applying Fubini’s theorem to the preceding equation, Eq.
(1.2) now follows easily from the following lemma.

######  Lemma 2.2.

For any $z\in\mathbb{C}^{n}$, the following formula holds

$c_{n,k}\int_{\mathbb{S}^{n}}\big{|}<z,\mu>\big{|}^{2k}d\sigma(\mu)=\big{\lVert}z\big{\rVert}_{2}^{2k}.$
(2.16)

###### Proof.

In view of the multinomial theorem, we can write

$\big{|}<z,\mu>\big{|}^{2k}=\Big{|}\big{(}<z,\mu>\big{)}^{k}\Big{|}^{2}=\Big{|}\sum_{\begin{subarray}{c}|\alpha|=k\\\
\alpha\in\mathbb{N}^{n}_{0}\end{subarray}}\binom{k}{\alpha}z^{\alpha}\
\overline{\mu}^{\alpha}\Big{|}^{2}=\sum_{\begin{subarray}{c}|\alpha|=|\beta|=k\\\
\alpha,\beta\in\mathbb{N}^{n}_{0}\end{subarray}}\binom{k}{\alpha}\binom{k}{\beta}z^{\alpha}\
\overline{z}^{\beta}\mu^{\beta}\ \overline{\mu}^{\alpha}.$

Integrating both sides of the preceding equality w.r.t. $\mu$ over
$\mathbb{S}^{n}$, and using Lemma 2.1 in [23] we get

$\int_{\mathbb{S}^{n}}\big{|}<z,\mu>\big{|}^{2k}d\sigma(\mu)=\sum_{\begin{subarray}{c}|\alpha|=k\\\
\alpha\in\mathbb{N}^{n}_{0}\end{subarray}}\binom{k}{\alpha}\binom{k}{\alpha}\dfrac{(n-1)!\alpha!}{(n-1+|\alpha|)!}z^{\alpha}\
\overline{z}^{\alpha}=\frac{k!(n-1)!}{(n-1+k)!}\sum_{\begin{subarray}{c}|\alpha|=k\\\
\alpha\in\mathbb{N}^{n}_{0}\end{subarray}}\binom{k}{\alpha}z^{\alpha}\
\overline{z}^{\alpha}=\frac{\big{\lVert}z\big{\rVert}_{2}^{2k}}{c_{n,k}}.$

∎

##  3\. Complete homogeneous symmetric polynomials and gauge functions

The main purpose of this section is to apply our results concerning the
unitarily invariant norms and that of Q-matrix norms to the study of the
q-complete homogeneous symmetric functions as introduced in [23].

Recall that in [23], we pointed out that for any $A\in M_{n}(\mathbb{C})$ Eq.
(1.1) can be reformulated as

$\int_{\mathbb{S}^{n}}\big{(}<A\xi,\xi>\big{)}^{k}d\sigma(\xi)=H_{k}(\lambda),$
(3.1)

where $\lambda=\lambda(A)$ and $H_{k}$ is the normalized complete
$k$-homogeneous polynomial in $n$ complex variables given by

$H_{k}(z)=\frac{k!(n-1)!}{(n+k-1)!}\sum_{\begin{subarray}{c}|\alpha|=k\\\
\alpha\in\mathbb{N}^{n}\end{subarray}}z^{\alpha},\quad z\in\mathbb{C}^{n}.$

In particular, for every $x\in\mathbb{R}^{n}$

$H_{k}(x)=\int_{\mathbb{S}^{n}}\big{(}<X\xi,\xi>\big{)}^{k}d\sigma(\xi),$
(3.2)

where $X:=\texttt{diag}(x_{1},\cdots,x_{n})$. Also, we need from [23] the
following case of the generalization for the restriction of $H_{k}$ onto
$\mathbb{R}^{n}_{+}$. Given $q\geq 1$ the $q$-homogeneous function $H_{q}$ is
defined $\mathbb{R}^{n}_{+}$ by

$H_{q}(x)=\int_{\mathbb{S}^{n}}\big{(}<X\xi,\xi>\big{)}^{q}d\sigma(\xi),$
(3.3)

Let us first introduce some more notation. Given $x,y\in\mathbb{R}^{n}$ and
$p\geq 0$, we write

$|x|=(|x_{1}|,\cdots,|x_{n}|),\quad|x|^{p}=(|x_{1}|^{p},\cdots,|x_{n}|^{p}),\quad
xy=(x_{1}y_{1},\cdots,x_{n}y_{n})\quad\mbox{ and}\quad
x^{\downarrow}=(x_{[1]},\cdots,x_{[n]}),$

where $x_{[i]}$ denotes the ith component obtained from the components of $x$
after rearranging them in decreasing order. We also use the usual weak
majorization notation i.e.

$x\prec_{w}y\quad\mbox{if}\quad\sum^{k}_{i=1}x_{[i]}\leq\sum^{j}_{i=1}y_{[i]},\
\ j=1\cdots n$

If $x\prec_{w}y$ and $\sum^{n}_{i=1}=x_{i}\leq\sum^{n}_{i=1}y_{i}$, we say
that $x$ is mjorized by $y$ and we write $x\preccurlyeq y$.

Motivated by our construction of the unitarily invariant norms and the Q-norms
obtained in Section 2, we now state our first result for this section.

######  Theorem 3.1.

For each $q\geq 1$, the map $\Phi_{q}$ defined on $\mathbb{R}^{n}$ by

$\Phi_{q}(x):=H_{q}^{\frac{1}{q}}(|x|),$

is a symmetric gauge function.

###### Proof.

By Theorem 2.1, we know that

$N_{q}(A)=\Bigg{(}\int_{\mathbb{S}^{n}}\Big{\lVert}|A|^{\frac{1}{2}}\xi\Big{\rVert}_{2}^{2q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}=\Bigg{(}\int_{\mathbb{S}^{n}}\big{(}<|A|\xi,\xi>\big{)}^{q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}$

defines a unitarily invariant norm on $M_{n}(\mathbb{C})$. Hence it defines a
s.g.f. (see, for example, [6]) denoted by $\Phi_{q}$, on $\mathbb{R}^{n}$
given by

$\Phi_{q}(x)=N_{q}(X),$

where $X:=\texttt{diag}(x_{1},\cdots,x_{n})$. From (3.3) we obtain

$N_{q}(X)=\Bigg{(}\int_{\mathbb{S}^{n}}\big{(}<|X|\xi,\xi>\big{)}^{q}d\sigma(\xi)\Bigg{)}^{\frac{1}{q}}=H^{\frac{1}{q}}_{q}(x).$

This shows that $\Phi_{q}=H^{\frac{1}{q}}_{q}$ and the proof is complete. ∎

It is worth mentioning here that even in the case where $q=k\in\mathbb{N}$,
and to the best of our knowledge, the connection of the analysis between the
homogeneous symmetric polynomials and the symmetric gauge functions has not
been pointed out before. However, one could obtain this result (for $q=k$)
from the literature by combining the facts that $H_{k}$ is $k$-homogeneous,
positive definite on $\mathbb{R}^{n}_{+}$ and invariant under permutation of
coordinates with the convexity results obtained by Marcus and Lopes [29] (cf.
pages 116 and 119 in[30]).

As a result, we have the following conclusion.

######  Corollary 3.1.

Given two real numbers $q\geq 1$ and $p\geq 1$. Then, for any
$x,y\in\mathbb{R}^{n}$ we have

$H_{q}(|xy|)\leq H^{\frac{1}{2}}_{q}(x^{2})H^{\frac{1}{2}}_{q}(y^{2})$ (3.4)

and

$H^{\frac{1}{qp}}_{q}(|x+y|^{p})\leq
H^{\frac{1}{qp}}_{k}(|x|^{p})+H^{\frac{1}{qp}}_{q}(|y|^{p}).$ (3.5)

###### Proof.

It suffices to see that the two inequalities are Hölder and Minkowski’s
inequality for symmetric gauge functions (cf. for example [6], p: 88) when
applied to $H^{\frac{1}{q}}_{q}$. ∎

Equation (3.4) is a generalization for the multiplicatively convex (cf. [34])
property for $H_{k}$ obtained by Y.-M Chu et al. in 2011 (cf. [11]). Moreover,
Eq. (3.5) is a generalization for Theorem 3.9 in [39], obtained by S. Sra in
2020, where a discrete version for fractional power of Minkowski’s inequality
was needed.

Taking advantage of the conclusion of Problem IV . 5 . 2 . in [6], we obtain
the following result which to the best of our knowledge is new even for the
case $q=k\in\mathbb{N}$.

######  Corollary 3.2.

Let $q\geq 1$ and $p\in]0,1[$. Then, for any $x,y\in\mathbb{R}^{n}$ we have

$H^{\frac{1}{pq}}_{q}(|x+y|^{p})\leq
2^{\frac{1}{p}-1}\Big{[}H^{\frac{1}{pq}}_{q}(|x|^{p})+H^{\frac{1}{pq}}_{q}(|y|^{p})\Big{]}.$
(3.6)

Recall that every s.g.f. $\Phi$ is (convex and monotone) on
$\mathbb{R}^{n}_{+}$ (cf. for example, [6], p. 45) hence it is strongly
isotone i.e. given $x,y\in\mathbb{R}^{n}_{+}$ then

$x\preccurlyeq_{w}y\Longrightarrow\Phi(x)\leq\Phi(y).$

Thus, we obtain the following.

######  Corollary 3.3.

Let $q\geq 1$ then for any $x,y\in\mathbb{R}^{n}_{+}$ we have

$x\preccurlyeq_{w}y\Longrightarrow H_{q}(x)\leq H_{q}(y).$ (3.7)

###### Proof.

From the previous discussion, we know that $H^{\frac{1}{q}}_{q}$ is strongly
isotone on $\mathbb{R}^{n}_{+}$. Therefore,

$x\preccurlyeq_{w}y\Longrightarrow H^{\frac{1}{q}}_{q}(x)\leq
H^{\frac{1}{q}}_{q}(y)\Longrightarrow H_{q}(x)\leq H_{q}(y),\quad
x,y\in\mathbb{R}^{n}_{+}.$

∎

It is worthy to note here that Eq. (3.7) is a generalization for the work of
Guan in [17]. Furthermore, motivated by the work of D. B. Hunter in [18], T.
Tao presented (in his blog) a proof for the Schur-convexity of the even degree
complete symmetric polynomials on $\mathbb{R}^{n}$. His proof was based on
applying some differential operators to $H_{2k}$.

Now in connection with this, we are in a position to present a simpler proof
for the Schur convexity property and to extend it to $\mathbb{R}^{n}$. For
this reason, we shall need a characterization for w.u.i. norms which was
obtained by Chi-K. Li and N. K. Tsing in [26].

######  Lemma 3.1 ([26], Theorem 4.1).

A map $N:\mathcal{H}_{n}\longrightarrow\mathbb{R}$ is (the restriction of) a
w.u.i norm if and only if there is a Schur convex norm
$\Phi:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ such that

$N(A)=\Phi(\lambda(A)),\quad A\in\mathcal{H}_{n}.$ (3.8)

Here is our result generalizing the Schur-convexity for $H_{2k}$ on
$\mathbb{R}^{n}$.

######  Corollary 3.4.

For each $q\geq 1$, consider the map $\mathcal{H}_{2q}$ defined on
$\mathbb{R}^{n}$ by

$\mathcal{H}_{2q}(x):=\bigg{[}\int_{\mathbb{S}^{n}}\big{(}<X\xi,\xi>\big{)}^{2}\bigg{]}^{q}d\sigma(\xi),$

where $X=\texttt{diag}(x_{1},\cdots,x_{n})$. Then, $\mathcal{H}_{2q}$ is Schur
convex on $\mathbb{R}^{n}$.

###### Proof.

From Theorem 2.1 in [10], we know that

$\Phi^{\prime}_{2q}(A)=\Big{(}\int_{\mathbb{S}^{n}}\Big{|}<A\xi,\xi>\Big{|}^{2q}d\sigma(\xi)\Big{)}^{\frac{1}{2q}}$

is a w.u.i. norm on $M_{n}(\mathbb{C})$. Given $x\in\mathbb{R}^{n}$ which
obviously means that $X\in\mathcal{H}$ so that

$\Phi^{\prime}_{2q}(X)=\Big{(}\int_{\mathbb{S}^{n}}\Big{|}<X\xi,\xi>\Big{|}^{2q}d\sigma(\xi)\Big{)}^{\frac{1}{2q}}=\Big{(}\int_{\mathbb{S}^{n}}\bigg{[}\big{(}<X\xi,\xi>\big{)}^{2}\bigg{]}^{q}d\sigma(\xi)\Big{)}^{\frac{1}{2q}}=\mathcal{H}^{\frac{1}{2q}}_{2q}(x).$

Applying the preceding lemma to $N=\Phi^{\prime}_{2q}$, we obtain that
$\mathcal{H}^{\frac{1}{2q}}_{2q}$ is a Schur convex (norm) i.e.

$x\prec
y\Longrightarrow\mathcal{H}^{\frac{1}{2q}}_{2q}(x)\leq\mathcal{H}_{2q}(y)\Longrightarrow\mathcal{H}_{2q}(x)\leq\mathcal{H}^{\frac{1}{2q}}_{2q}(y),\quad
x,y\in\mathbb{R}^{n}.$

∎

We conclude this section by providing an interpolating sequence of
inequalities for $H_{q}$ on $\mathbb{R}^{n}_{+}$ as follows.

######  Theorem 3.2.

For any pair of real numbers $q\geq p\geq 1$, the following formula holds

$H_{q}(x^{p})\leq H_{p}(x^{q}),\quad x\in\mathbb{R}^{n}_{+}.$

###### Proof.

From McCarty inequality, we know that

$\Big{(}\big{<}X^{p}\xi,\xi\big{>}\Big{)}^{\frac{q}{p}}\leq\big{<}(X^{p})^{\frac{q}{p}}\xi,\xi\big{>}=\big{<}X^{q}\xi,\xi\big{>},$

for each $\xi\in\mathbb{S}^{n}$. The result now follows easily by taking the
$p$th power for both sides of the above inequality and then integrating. ∎

######  Remark 3.1.

One may use Theorem 3.1 to obtain other generalization for similar
inequalities involving $q$-homogeneous symmetric functions. We shall point out
a particular application in this direction. Following [9], given two s.g.f.
$f$ and $g$ one can construct a new s.g.f. $f\div g$ defined on the set of
decreasing tuples $\mathbb{R}^{n}_{\downarrow}$ by

$f\div g(x):=\max\Big{\\{}\frac{f(yx)}{g(y)}\ \ :\
y\in\mathbb{R}^{n}_{\downarrow}\backslash\\{0\\}\Big{\\}}$

and then extend it by symmetry to $\mathbb{R}^{n}$. Now let
$q,q^{\prime}\in[1,\infty[$ and consider $f:=H_{q}^{\frac{1}{q}}$ and
$g:=H_{q^{\prime}}^{\frac{1}{q^{\prime}}}$ then for
$x\in\mathbb{R}^{n}_{\downarrow}$ the function

$H_{q,q^{\prime}}(x):=\max\Big{\\{}\frac{H_{q}^{\frac{1}{q}}\big{(}|y||x|\big{)}}{H_{q^{\prime}}^{\frac{1}{q^{\prime}}}\big{(}|y|\big{)}}\
\ :\ y\in\mathbb{R}^{n}_{\downarrow}\backslash\\{0\\}\Big{\\}}$

extends to a s.g.f. on $\mathbb{R}^{n}$. In particular, if $q=q^{\prime}$ then
by (3.4) it follows that $H_{q,q^{\prime}}=H^{\frac{1}{q}}_{q}$. As
$H_{q,q^{\prime}}$ is a s.g.f. then such generalization for $q$-homogeneous
functions satisfies Equations (3.4)-(3.7).

##  4\. Explicit value for a class of w.u.i. norms induced by $L^{k}$-norms

This section is devoted to continue the work of R. Bhatia and J.R. Holbrook in
[10] concerning giving expressions for the w.u.i. norms (on
$M_{n}(\mathbb{C})$) associated to the natural $L^{p}$ norms on
$C(\mathbb{S}^{n})$. In the next theorem, we combine some of the results
obtained in [10] to suit our needs.

######  Theorem 4.1 ([10]).

Let $A,B\in M_{n}(\mathbb{C}))$ then

$\int_{\mathbb{S}^{n}}<A\xi,\xi><B\xi,\xi>d\sigma(\xi)=\frac{1}{c_{n,2}}\Big{(}tr(AB)+tr(A)tr(B)\Big{)}.$
(4.1)

In particular,

$\Phi^{\prime
2}_{2}(A):=\int_{\mathbb{S}^{n}}\Big{|}<A\xi,\xi>\Big{|}^{2}d\sigma(\xi)=\frac{\lVert
A\rVert_{F}^{2}+|tr(A)|^{2}}{c_{n,2}}.$ (4.2)

Moreover, if $E$ denotes the expectation and $V$ is the variance (w.r.t.
$d\sigma$) then the Frobenius norm satisfies

$\big{\lVert}A\big{\rVert}^{2}_{(2)}=nE\big{(}|f_{A}|^{2}\big{)}+n^{2}V(f_{A}),$
(4.3)

where $f_{A}$ is defined by (1.6).

The authors proved (4.1) using some techniques that are based on Riesz
representation theorem that deals with characterizing all w.u.i. sesquilinear
forms on $M_{n}(\mathbb{C})$. Taking $B=I_{n}$ in (4.1), we clearly obtain Eq.
(1.1) for $k=1$. Choosing $B=A^{\star}$, (Eq. 1.1) follows easily and finally
using (4.2 and (1.1) Eq. (4.3) is obtained. The argument on the sesquilinear
forms which is used in [10] is not sufficient to examine the explicit value of

$\Phi^{\prime}_{k}(A):=\Big{(}\int_{\mathbb{S}^{n}}\Big{|}<A\xi,\xi>\Big{|}^{k}d\sigma(\xi)\Big{)}^{\frac{1}{k}},\
\ A\in M_{n}(\mathbb{C}),$ (4.4)

for general $k\in\mathbb{N}$ (cf. Section IV. in [6] for a motivation to this
problem). So far, we have obtained the following results: []

$\Phi^{\prime}_{k}(A)=\Big{[}Tr(\vee^{k}A)\Big{]}^{\frac{1}{k}}=N_{k}(A)=\Big{[}\frac{1}{c_{n,k}}\sum_{l=1}^{k}\sum_{\beta\in
S^{k}_{l}}\frac{1}{z_{\beta}}\prod_{t=1}^{l}\big{\lVert}A\big{\rVert}^{\beta_{t}}_{{}_{(\beta_{t})}}\Big{]}^{\frac{1}{k}},\quad
A\in\mathcal{H}_{n}^{+}$ (4.5)
$\Phi^{\prime}_{2k}(A)=\Big{[}Tr(\vee^{2k}A)\Big{]}^{\frac{1}{2k}}=N_{2k}(A)=\Big{[}\frac{1}{c_{n,2k}}\sum_{l=1}^{2k}\sum_{\beta\in
S^{2k}_{l}}\frac{1}{z_{\beta}}\prod_{t=1}^{l}\big{\lVert}A\big{\rVert}^{\beta_{t}}_{{}_{(\beta_{t})}}\Big{]}^{\frac{1}{2k}},\quad
A\in\mathcal{H}_{n}.$ (4.6)

Our next result deals with presenting an upper estimation for
$\Phi^{\prime}_{2p}$ using the Q-norms defined by (2.15)

######  Proposition 4.1.

Let $A\in M_{n}(\mathbb{C})$ and let $p\geq 1$ then

$\Phi^{\prime 2p}_{2p}(A)\leq\frac{1}{2}\Big{(}\Phi^{\prime
p}_{p}(A^{2})+N^{\prime 2p}_{p}(A)\Big{)}.$ (4.7)

###### Proof.

Recall that for any $A\in M_{n}(\mathbb{C})$ we have

$\Big{|}<A\xi,\xi>\Big{|}^{2p}\leq\frac{1}{2}\Bigg{(}\Big{|}<A^{2}\xi,\xi>\Big{|}^{p}+\Big{\lVert}A\xi\Big{\rVert}^{p}\Big{\lVert}A^{\star}\xi\Big{\rVert}^{p}\Bigg{)}.$

Integrating both sides of the preceding inequality leads to

$\displaystyle\Phi^{\prime 2p}_{2p}(A)$
$\displaystyle\leq\frac{1}{2}\Big{(}\Phi^{\prime
p}_{p}(A^{2})+\int_{\mathbb{S}^{n}}\Big{\lVert}A\xi\Big{\rVert}^{p}\Big{\lVert}A^{\star}\xi\Big{\rVert}^{p}d\sigma(\xi)\Big{)}$
$\displaystyle\leq\frac{1}{2}\Bigg{(}\Phi^{\prime
p}_{p}(A^{2})+\Big{[}\int_{\mathbb{S}^{n}}\Big{\lVert}A\xi\Big{\rVert}^{2p}d\sigma(\xi)\Big{]}^{\frac{1}{2}}\Big{[}\int_{\mathbb{S}^{n}}\Big{\lVert}A^{\star}\xi\Big{\rVert}^{2p}d\sigma(\xi)\Big{]}^{\frac{1}{2}}\Bigg{)}$
(4.8) $\displaystyle=\frac{1}{2}\Big{(}\Phi^{\prime p}_{p}(A^{2})+N^{\prime
2p}_{p}(A)\Big{)},$ (4.9)

where (4.8) is the usual Hölder inequality and (4.9) follows from the fact
that $N^{\prime}_{p}$ is unitarly invariant. ∎

The next theorem is a generalization of (4.1). Its proof uses Lemma A.1, which
for convenience, is given in the appendix below.

######  Theorem 4.2.

Let $A,B,C,D\in M_{n}(\mathbb{C})$ be arbitrary matrices and put
$c_{n}:=n(n+1)(n+2)(n+3)$. Consider $\sigma$ the normalized probability
measure on the unit sphere $\mathbb{S}^{n}$ of $\mathbb{C}^{n}$. Then, [] []

$\displaystyle
c_{n}\int_{\mathbb{S}^{n}}<A\xi,\xi><B\xi,\xi><C\xi,\xi><D\xi,\xi>d\sigma(\xi)=$
$\displaystyle
tr\Big{[}ABCD+ABDC+ACBD+ACDB+ADBC+ADCB\Big{]}+tr(A)tr\Big{[}BCD+BDC\Big{]}$
$\displaystyle+tr(B)tr\Big{[}ACD+ADC\Big{]}+tr(C)tr\Big{[}ABD+ADB\Big{]}+tr(D)tr\Big{[}ABC+ACB\Big{]}$
$\displaystyle+tr(AB)\Big{[}tr(CD)+tr(C)tr(D)\Big{]}+tr(AC)\Big{[}tr(BD)+tr(B)tr(D)\Big{]}$
$\displaystyle+tr(AD)\Big{[}tr(BC)+tr(B)tr(C)\Big{]}+tr(A)\Big{[}tr(B)tr(CD)+tr(C)tr(BD)+tr(D)tr(BC)\Big{]}$
$\displaystyle+tr(A)tr(B)tr(C)tr(D).$ (4.10)

###### Proof.

From (1.1), we know that:

$c_{n}\int_{\mathbb{S}^{n}}\big{(}<A\xi,\xi>\big{)}^{4}d\sigma(\xi)=6tr(A^{4})+8[tr(A^{3})].tr(A)+3[tr(A^{2})]^{2}+6[tr(A^{2})](trA)^{2}+[tr(A)]^{4}.$
(4.11)

Let $x,y,z$ be formal variables and put

$P(x,y,z)=A+xB+yC+zD\in M_{n}(\mathbb{C}[x,y,z]).$ (4.12)

Applying (4.11) to $P(x,y,z)$ yields the following equality for multinomials
in $\mathbb{C}[x,y,z]$

$\displaystyle
c_{n}\int_{\mathbb{S}^{n}}\big{(}<P(x,y,z)\xi,\xi>\big{)}^{4}d\sigma(\xi)=$
$\displaystyle
6tr\big{[}P^{4}(x,y,z)\big{]}+8tr\big{[}P^{3}(x,y,z)\big{]}.tr\big{[}P(x,y,z)\big{]}$
$\displaystyle+3\bigg{[}tr\big{[}P^{2}(x,y,z)\big{]}\bigg{]}^{2}+6tr\big{[}P^{2}(x,y,z)\big{]}\cdot\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{2}$
$\displaystyle+\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{4}.$ (4.13)

Applying the multinomial theorem to the integrand in the above equality yields
[]

$\begin{split}&\int_{\mathbb{S}^{n}}\Big{(}<P(x,y,z)\xi,\xi>\Big{)}^{4}d\sigma(\xi)\\\
&=\int_{\mathbb{S}^{n}}\Bigg{(}<A\xi,\xi>+x<B\xi,\xi>+y<C\xi,\xi>+z<D\xi,\xi>\Bigg{)}^{4}d\sigma(\xi)\\\
&=\int_{\mathbb{S}^{n}}\sum_{\begin{subarray}{c}|\alpha|=4\\\
\alpha\in\mathbb{N}_{0}^{4}\end{subarray}}\binom{4}{\alpha}\Big{(}<A\xi,\xi>,x<B\xi,\xi>,y<C\xi,\xi>,z<D\xi,\xi>\Big{)}^{\alpha}d\sigma(\xi)\\\
&=\sum_{\begin{subarray}{c}\alpha_{1}+\cdots+\alpha_{4}=4\\\
\alpha\in\mathbb{N}_{0}^{4}\end{subarray}}\binom{4}{\alpha}\int_{\mathbb{S}^{n}}\big{(}<A\xi,\xi>\big{)}^{\alpha_{1}}\big{(}x<B\xi,\xi>\big{)}^{\alpha_{2}}\big{(}y<C\xi,\xi>\big{)}^{\alpha_{3}}\big{(}z<D\xi,\xi>\big{)}^{\alpha_{4}}d\sigma(\xi).\end{split}$

In particular, the $xyz$-variable corresponds to $\alpha=(1,1,1,1)$ i.e. the
coefficient of the $xyz$-variable on the left side of (4) is given by

$24c_{n}\int_{\mathbb{S}^{n}}<A\xi,\xi><B\xi,\xi><C\xi,\xi><D\xi,\xi>d\sigma(\xi).$
(4.14)

The coefficient of the $xyz$-variable on the right hand side of (4) is given
by 24 multiplied by the right hand side of (4.2) and then the theorem follows
form Lemma A.1. ∎

The following corollaries are direct applications of the preceding theorem.
Recalling that $\Re(.)$ denotes the real part and setting first $D=I_{n}$ in
(4.2) we obtain the following conclusion.

######  Corollary 4.1.

Let $A,B,C\in M_{n}(\mathbb{C})$ and put $d_{n}:=6c_{n,3}=n(n+1)(n+2)$ then

$\displaystyle
d_{n}\int_{\mathbb{S}^{n}}<A\xi,\xi><B\xi,\xi><C\xi,\xi>d\sigma(\xi)=tr$
$\displaystyle\Big{[}ABC+ACB\Big{]}+tr(A)tr(BC)+tr(B)tr(AC)$
$\displaystyle+tr(C)tr(AB)+tr(A)tr(B)tr(C).$ (4.15)

In particular, if $C\in\mathcal{H}_{n}^{+}$ and
$d\mu(\xi):=<C\xi,\xi>d\sigma(\xi)$ then

$d_{n}\int_{\mathbb{S}^{n}}\Big{|}<A\xi,\xi>\Big{|}^{2}d\mu(\xi)=tr\Big{[}AA^{\star}C+ACA^{\star}\Big{]}+2\Re\Big{[}tr(A)tr(A^{\star}C)\Big{]}+tr(C)\lVert
A\rVert^{2}_{F}+\Big{|}tr(A)\Big{|}^{2}tr(C).$ (4.16)

Note that the preceding corollary generalizes (4.1) and(4.2). Now let $B=A$
and $C=D=A^{\star}$ this time, we get

######  Corollary 4.2.

Let $A\in M_{n}(\mathbb{C})$ and $c_{n}:=n(n+1)(n+2)(n+3)$, then

$\displaystyle c_{n}\int_{\mathbb{S}^{n}}\Big{|}<Au,u>\Big{|}^{4}d\sigma(u)$
$\displaystyle=4tr\big{(}A^{2}A^{\star
2}\big{)}+2tr\Big{[}\big{(}A^{\star}A\big{)}^{2}\Big{]}+2\Big{[}tr\big{(}A^{\star}A\big{)}\Big{]}^{2}+8\Re\Big{[}tr\big{(}A\big{)}tr\big{(}AA^{\star
2}\big{)}\Big{]}$
$\displaystyle+2\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}+4\Big{[}tr\big{(}A^{\star}A\big{)}\Big{]}\cdot\Big{|}tr(A)\Big{|}^{2}+\Big{|}tr(A)\Big{|}^{4}.$

Equivalently $\Phi^{\prime}_{4}(A)$ is given by

$\left(\frac{4\big{\lVert}A^{2}\big{\rVert}_{F}^{2}+2\big{\lVert}A\big{\rVert}^{4}_{(4)}+2\big{\lVert}A\big{\rVert}_{F}^{4}+8\Re\Big{[}tr\big{(}A\big{)}tr\big{(}AA^{\star
2}\big{)}\Big{]}+2\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+4\big{\lVert}A\big{\rVert}_{F}^{2}\Big{|}tr(A)\Big{|}^{2}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}+\Big{|}tr(A)\Big{|}^{4}}{c_{n}}\right)^{\frac{1}{4}}.$

(4.17)

######  Remark 4.1.

In the connection to the study for estimations of the w.u.i. norms associated
to the $L^{p}$-norms on $C(\mathbb{S}^{n})$, one can find an upper estimation
for $\Phi^{\prime}_{p}$, whenever $p\in]2,4[$ in-terms $\Phi^{\prime}_{2}$ and
$\Phi^{\prime}_{4}$. By a direct application of Riesz-Thorin interpolation
theorem (cf. Theorem 2.5 in[41]) we obtain

$\Phi^{\prime}_{p}(A)\leq\big{(}\Phi^{\prime}_{2}(A)\big{)}^{\frac{4}{p}-1}\big{(}\Phi^{\prime}_{4}(A)\big{)}^{2-\frac{4}{p}},\quad\mbox{for
any}\ A\in M_{n}(\mathbb{C}),$

where $\Phi^{\prime}_{2}$ and $\Phi^{\prime}_{4}$ are given (4.2) and (4.17),
respectively.

The expression for $\Phi^{\prime}_{4}$ involves many terms, even if we
restrict our attention to normal matrices. This motivates us to slightly
modify the $\Phi^{\prime}_{4}$-norm in order to obtain some (other) weakly
unitarily invariant norms with less numbers of terms in (4.17). This will also
lead to some probabilistic interpretation for a certain combination of the
Schatten 2 and 4-norms. Let $T:C(\mathbb{S}^{n})\longrightarrow
C(\mathbb{S}^{n})$ be the linear map defined by

$T(f)=f-\int_{\mathbb{S}^{n}}f.$ (4.18)

In addition, let $\varPsi_{1}$ and $\varPsi_{2}$ be two unitarily invariant
norms on $C(\mathbb{S}^{n})$ and consider

$\varPsi(f)=\Big{(}\varPsi_{1}^{4}(f)+\varPsi_{2}^{4}(Tf)\Big{)}^{\frac{1}{4}},\quad
f\in C(\mathbb{S}^{n}).$ (4.19)

Notice that, $T(f\circ U)=(Tf)\circ U$ for every $f\in C(\mathbb{S}^{n})$ and
each $U\in U_{n}.$ Therefore, $\varPsi$ is a unitarily invariant norm on
$C(\mathbb{S}^{n})$.

######  Theorem 4.3.

Let $\varPsi$ be the unitarily invariant norm on $C(\mathbb{S}^{n})$ defined
by (4.19), where

$\varPsi_{1}(\cdot)=\sqrt[4]{3d_{n}}\Big{\lVert}\cdot\Big{\rVert}_{L^{4}}\quad\mbox{and}\quad\varPsi_{2}(\cdot)=\sqrt[4]{nd_{n}}\Big{\lVert}\cdot\Big{\rVert}_{L^{4}}.$

Then the weakly unitarily invariant norm on $M_{n}(\mathbb{C})$ induced by
$\varPsi$, denoted by $N_{\varPsi}$, satisfies

$\displaystyle N^{4}_{\varPsi}(A)=$ $\displaystyle 4\lVert
A^{2}\rVert_{F}^{2}+2\lVert A\rVert^{4}_{(4)}+2\lVert
A\rVert_{F}^{4}+\frac{4}{n}\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+\frac{8}{n}\lVert
A\rVert_{F}^{2}\Big{|}tr(A)\Big{|}^{2}$
$\displaystyle+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}+\frac{3n-6}{n^{2}}\Big{|}tr(A)\Big{|}^{4}.$
(4.20)

###### Proof.

Let $\gamma=\frac{-tr(A)}{n}$ and put $f=f_{A}$. Integrating the identity,
$|f+\gamma|^{4}=|f|^{4}+|\gamma|^{4}+4|\gamma|^{2}|f|^{2}+2\Re\overline{\gamma}^{2}f^{2}+4(|f|^{2}+|\gamma|^{2})\Re\overline{\gamma}f$,
leads to []

$\displaystyle\int_{\mathbb{S}^{n}}|f+\gamma|^{4}=$
$\displaystyle\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+|\gamma|^{4}+4|\gamma|^{2}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}+\frac{2}{n(n+1)}\Re\Big{[}\overline{\gamma}^{2}\big{(}tr(A^{2})+n^{2}\gamma^{2}\big{)}\Big{]}$
$\displaystyle+4\Re\Big{[}\overline{\gamma}\big{(}\int_{\mathbb{S}^{n}}|f|^{2}f+|\gamma|^{2}\int_{\mathbb{S}^{n}}f\big{)}\Big{]}$
(4.21)
$\displaystyle=\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+|\gamma|^{4}+4|\gamma|^{2}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}+\frac{2}{n(n+1)}\Big{[}n^{2}|\gamma|^{4}+\Re\big{(}\overline{\gamma}^{2}tr(A^{2})\big{)}\Big{]}-4|\gamma|^{4}$
$\displaystyle+\frac{4}{d_{n}}\Re\Bigg{[}\overline{\gamma}\Big{(}2tr(A^{2}A^{\star})+2tr(A)tr(A^{\star}A)+tr(A^{\star})tr(A^{2})+\Big{|}tr(A)\Big{|}^{2}tr(A)\Big{)}\Bigg{]}$
(4.22)
$\displaystyle=\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+|\gamma|^{4}+4|\gamma|^{2}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}+\frac{2}{n(n+1)}\Big{[}n^{2}|\gamma|^{4}+\Re\big{(}\overline{\gamma}^{2}tr(A^{2})\big{)}\Big{]}-4|\gamma|^{4}$
$\displaystyle+\frac{4}{d_{n}}\Bigg{[}\Re\Big{(}2\overline{\gamma}tr(A^{2}A^{\star})-n\overline{\gamma}^{2}tr(A^{2})\Big{)}-2n|\gamma|^{2}\lVert
A\rVert_{F}^{2}-n^{3}|\gamma|^{4}\Bigg{]}$
$\displaystyle=\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+4|\gamma|^{2}\Big{[}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}-\frac{2n}{d_{n}}\lVert
A\rVert_{F}^{2}\Big{]}+\Big{(}\frac{2}{n(n+1)}-\frac{4n}{d_{n}}\Big{)}\Re\overline{\gamma}^{2}tr(A^{2})$
$\displaystyle+\frac{8}{d_{n}}\Re\Big{(}\overline{\gamma}tr(A^{2}A^{\star})\Big{)}+(-3+\frac{2n}{n+1}-\frac{4n^{3}}{d_{n}})|\gamma|^{4}$
$\displaystyle=\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+4|\gamma|^{2}\Big{[}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}-\frac{2n}{d_{n}}\lVert
A\rVert_{F}^{2}\Big{]}+\frac{4-2n}{d_{n}}\Re\overline{\gamma}^{2}tr(A^{2})+\frac{8}{d_{n}}\Re\Big{(}\overline{\gamma}tr(A^{2}A^{\star})\Big{)}$
$\displaystyle-\frac{5n^{3}+5n^{2}+6n}{d_{n}}|\gamma|^{4}.$

where (4.21) follows from (4.1) and (4.2), whereas (4.22) follows from (4.1).
Using the preceding equality, one gets
$\varPsi_{2}^{4}(Tf)=nd_{n}\int_{\mathbb{S}^{n}}|f+\gamma|^{4}$ is given by []

$nd_{n}\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+(8n-4n^{2})|\gamma|^{2}\lVert
A\rVert_{F}^{2}+(4n-2n^{2})\Re\overline{\gamma}^{2}tr(A^{2})+8n\Re\Big{(}\overline{\gamma}tr(A^{2}A^{\star})\Big{)}+(-n^{4}+3n^{3}-6n^{2})|\gamma|^{4}.$

By Eq. (4.17), we obtain that
$N^{4}_{\varPsi}(A)=3d_{n}\int_{\mathbb{S}^{n}}\Big{|}f\Big{|}^{4}+nd_{n}\int_{\mathbb{S}^{n}}|f+\gamma|^{4}$
is given by

$\displaystyle N^{4}_{\varPsi}(A)=$ $\displaystyle
c_{n}\big{(}\Phi^{\prime}_{4}(A)\big{)}^{4}+(8n-4n^{2})|\gamma|^{2}\lVert
A\rVert_{F}^{2}+(4n-2n^{2})\Re\overline{\gamma}^{2}tr(A^{2})+8n\Re\Big{(}\overline{\gamma}tr(A^{2}A^{\star})\Big{)}$
$\displaystyle+(-n^{4}+3n^{3}-6n^{2})|\gamma|^{4}$ $\displaystyle=$
$\displaystyle 4\lVert A^{2}\rVert_{F}^{2}+2\lVert A\rVert^{4}_{(4)}+2\lVert
A\rVert_{F}^{4}+8\Re\Big{[}tr\big{(}A\big{)}tr\big{(}AA^{\star
2}\big{)}\Big{]}+2\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}$
$\displaystyle+4\lVert
A\rVert_{F}^{2}\Big{|}tr(A)\Big{|}^{2}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}+\Big{|}tr(A)\Big{|}^{4}+(\frac{8}{n}-4)\Big{|}tr(A)\Big{|}^{2}\lVert
A\rVert_{F}^{2}+(\frac{4}{n}-2)\Re\big{[}trA^{\star}\big{]}^{2}tr(A^{2})$
$\displaystyle-8\Re\Big{(}tr\big{(}A^{\star}\big{)}tr(A^{2}A^{\star})\Big{)}+(-1+\frac{3}{n}-\frac{6}{n^{2}})\Big{|}tr(A)\Big{|}^{4}$
$\displaystyle=$ $\displaystyle 4\lVert A^{2}\rVert_{F}^{2}+2\lVert
A\rVert^{4}_{(4)}+2\lVert
A\rVert_{F}^{4}+\frac{4}{n}\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+\frac{8}{n}\lVert
A\rVert_{F}^{2}\Big{|}tr(A)\Big{|}^{2}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}$
$\displaystyle+\frac{3n-6}{n^{2}}\Big{|}tr(A)\Big{|}^{4}.$

∎

A similar method to the above technique yields a reduction for the terms in
(4.3). Indeed, it is enough to consider the following unitarily invariant norm
on $C(\mathbb{S}^{n})$

$\varPsi_{0}(f)=\big{(}\varPsi^{4}(f)+\varPsi_{3}^{4}(Tf)\big{)}^{\frac{1}{4}},\quad
f\in C(\mathbb{S}^{n}),$ (4.23)

where $\varPsi$ is the norm considered in the previous theorem and
$\varPsi_{3}(\cdot):=\sqrt{2n(n+1)}\Big{\lVert}\cdot\Big{\rVert}_{L^{2}(C(\mathbb{S}^{n}))}$.

######  Theorem 4.4.

Let $\varPsi_{0}$ be the unitarily invariant norm on $C(\mathbb{S}^{n})$
defined by (4.23). The weakly unitarily invariant norm on $M_{n}(\mathbb{C})$
induced by $\varPsi_{0}$, denoted by $N_{\varPsi_{0}}$, satisfies

$N^{4}_{\varPsi_{0}}(A)=4\lVert A^{2}\rVert_{F}^{2}+2\lVert
A\rVert^{4}_{(4)}+6\lVert
A\rVert_{F}^{4}+\frac{4}{n}\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}+\frac{3n-2}{n^{2}}\Big{|}tr(A)\Big{|}^{4}.$
(4.24)

###### Proof.

Let $f$ be as in the previous proof, then

$\displaystyle\Bigg{[}\int_{\mathbb{S}^{n}}\Big{|}f-\int_{\mathbb{S}^{n}}f\Big{|}^{2}\Bigg{]}^{2}$
$\displaystyle=\Bigg{[}\int_{\mathbb{S}^{n}}|f|^{2}-\Big{|}\int_{\mathbb{S}^{n}}f\Big{|}^{2}\Bigg{]}^{2}=\Bigg{[}\big{(}\Phi^{\prime}_{2}(A)\big{)}^{2}-\frac{\Big{|}tr\big{(}A\big{)}\Big{|}^{2}}{n^{2}}\Bigg{]}^{2}$
$\displaystyle=\Bigg{[}\dfrac{n\lVert
A\rVert_{F}^{2}-\Big{|}tr\big{(}A\big{)}\Big{|}^{2}}{n^{2}(n+1)}\Bigg{]}^{2}=\frac{1}{n^{4}(n+1)^{2}}\Bigg{[}n^{2}\lVert
A\rVert_{F}^{4}-2n\lVert
A\rVert_{F}^{2}\Big{|}tr\big{(}A\big{)}\Big{|}^{2}+\Big{|}tr\big{(}A\big{)}\Big{|}^{4}\Bigg{]}.$

Hence,

$\varPsi_{3}^{4}(Tf)=4n^{2}(n+1)^{2}\Bigg{[}\int_{\mathbb{S}^{n}}\Big{|}f-\int_{\mathbb{S}^{n}}f\Big{|}^{2}\Bigg{]}^{2}=4\lVert
A\rVert_{F}^{4}-\frac{8}{n}\lVert
A\rVert_{F}^{2}\Big{|}tr\big{(}A\big{)}\Big{|}^{2}+\frac{4}{n^{2}}\Big{|}tr\big{(}A\big{)}\Big{|}^{4}$

and (4.24) follows by adding the preceding equation to ((4.3)). ∎

The preceding two theorems can be combined together to provide another weakly
unitarily invariant norm. We state this result in a probabilistic language.

######  Corollary 4.3.

Given $A\in M_{n}(\mathbb{C})$, consider the random variable
$f_{A}(\xi)=<A\xi,\xi>$ on $(\mathbb{S}^{n},d\sigma)$. For any $\alpha\geq 0$,
the following identity holds

$\displaystyle 3d_{n}E(|f_{A}|^{4})+nd_{n}\mu_{4}^{4}(f_{A})+4\alpha
n^{2}(n+1)^{2}V^{2}(f_{A})=$ $\displaystyle 4\lVert
A^{2}\rVert_{F}^{2}+2\lVert A\rVert^{4}_{(4)}+(2+4\alpha)\lVert
A\rVert_{F}^{4}+\frac{4}{n}\Re\Bigg{[}\big{[}tr(A^{2})\big{]}\big{[}trA^{\star}\big{]}^{2}\Bigg{]}+\frac{8}{n}(1-\alpha)\lVert
A\rVert_{F}^{2}\Big{|}tr(A)\Big{|}^{2}+\Big{|}tr\big{(}A^{2}\big{)}\Big{|}^{2}$
$\displaystyle+\frac{3n-6+4\alpha}{n^{2}}\Big{|}tr(A)\Big{|}^{4},$

where $E$ is the expectation, $V$ is the variance and $\mu_{4}$ is the fourth
central moment.

Motivated by the results of this section, we provide a method for finding an
expression of $\Phi^{\prime}_{2k}$ on $M_{n}(\mathbb{C})$ and for arbitrary
$k\in\mathbb{N}$. we shall first start by exploring a method to find an
expression of

$\int_{\mathbb{S}^{n}}\prod_{i=1}^{k}<A_{i}\xi,\xi>d\sigma(\xi),$ (4.25)

for arbitrary $A_{1},\cdots,A_{k}\in M_{n}(\mathbb{C})$. In order to do that,
let us consider one of the representations obtained in Section 2 say

$\int_{\mathbb{S}^{n}}\big{(}<A\xi,\xi>\big{)}^{k}d\sigma(\xi)=Tr(\vee^{k}A).$
(4.26)

Let $x_{2},x_{3},\cdots,x_{k}$ be formal variables and for convenience put
$x_{1}=1$, then replace $A$ in the preceding equation by the following matrix

$B=A_{1}+x_{2}A_{2}+x_{3}A_{3}+\cdots+x_{k}A_{k}.$

which has multinomial entries, we obtain

$\displaystyle\int_{\mathbb{S}^{n}}\big{(}<B\xi,\xi>\big{)}^{k}d\sigma(\xi)$
$\displaystyle=\int_{\mathbb{S}^{n}}\Big{(}\sum_{i=1}^{k}x_{i}<A_{i}\xi,\xi>\Big{)}^{k}d\sigma(\xi)$
$\displaystyle=\sum_{\begin{subarray}{c}|\alpha|=k\\\
\alpha\in\mathbb{N}_{0}^{k}\end{subarray}}\binom{k}{\alpha}x^{\alpha}\int_{\mathbb{S}^{n}}\prod_{i=1}^{k}(<A_{i}\xi,\xi>)^{\alpha_{i}}d\sigma(\xi)$
(4.27)
$\displaystyle=Tr\Bigg{(}\vee^{k}\Big{(}\sum_{i=1}^{k}x_{i}A_{i}\Big{)}\Bigg{)}.$
(4.28)

The coefficient of $x_{2}x_{3}...x_{k}$ in (4.27) is given by
$k!\int_{\mathbb{S}^{n}}\prod_{i=1}^{k}<A_{i}\xi,\xi>d\sigma(\xi)$.
Identifying this with the coefficient of $x_{2}x_{3}...x_{k}$ in (4.28)
provides an explicit value for (4.25). Notice that if the explicit value of
(4.25) is given for a certain $k_{0}$, then one can directly obtain (4.25) for
all $k\in{1,2,\cdots,k_{0}}$. Moreover, given the explicit value of (4.25) at
$l=2k$ then choosing $A_{1}=A_{2}=\cdots=A_{k}=A$ and
$A_{k+1}=A_{k+2}=\cdots=A_{2k}=A^{\star}$ provides the exact value of
$\Phi^{\prime}_{2k}(A)$.

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##  Appendix A Appendix

The appendix is devoted to find the explicit coefficient for the
$xyz$-variable in the multinomial (4).

######  Lemma A.1.

The coefficent of the $xyz$-variable in

$\begin{split}6tr\big{[}P^{4}(x,y,z)\big{]}&+8tr\big{[}P^{3}(x,y,z)\big{]}.tr\big{[}P(x,y,z)\big{]}+3\bigg{[}tr\big{[}P^{2}(x,y,z)\big{]}\bigg{]}^{2}\\\
&+6tr\big{[}P^{2}(x,y,z)\big{]}\cdot\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{2}+\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{4}.\end{split}$
(A.1)

is given by []

$\begin{split}24\Bigg{\\{}tr&\Big{[}ABCD+ABDC+ACBD+ACDB+ADBC+ADCB\Big{]}+tr(A)tr\Big{[}BCD+BDC\Big{]}\\\
&+tr(B)tr\Big{[}ACD+ADC\Big{]}+tr(C)tr\Big{[}ABD+ADB\Big{]}+tr(D)tr\Big{[}ABC+ACB\Big{]}\\\
&+tr(AB)\Big{[}tr(CD)+tr(C)tr(D)\Big{]}+tr(AC)\Big{[}tr(BD)+tr(B)tr(D)\Big{]}\\\
&+tr(AD)\Big{[}tr(BC)+tr(B)tr(C)\Big{]}+tr(A)\Big{[}tr(B)tr(CD)+tr(C)tr(BD)+tr(D)tr(BC)\Big{]}\\\
&+tr(A)tr(B)tr(C)tr(D)\Bigg{\\}}.\end{split}$ (A.2)

The proof is achieved by computing the exact coefficient for the
$xyz$-variable in each of the five terms of (A.1) and this is presented in the
next propositions.

######  Proposition A.1.

The coefficient of $xyz$ in $P^{4}(x,y,z)$ is given by

$\displaystyle(A$ $\displaystyle B+BA)(CD+DC)+(AC+CA)(BD+DB)+(AD+DA)(BC+CB)$
$\displaystyle+(BC+CB)(AD+DA)+(BD+DB)(AC+CA)+(CD+DC)(AB+BA).$ (A.3)

In particular, the $xyz$-coefficient of $tr\big{[}P^{4}(x,y,z)\big{]}$ is
given by

$4tr\Big{[}ABCD+ABDC+ACBD+ACDB+ADBC+ADCB\Big{]}.$ (A.4)

###### Proof.

The matrix $P^{2}(x,y,z)$ is given by

$\displaystyle P^{2}(x,y,z)$
$\displaystyle=\big{(}A+xB+yC+zD\big{)}\big{(}A+xB+yC+zD\big{)}$
$\displaystyle=A^{2}+(AB+BA)x+(AC+CA)y+(AD+DA)z+(BC+CB)xy$
$\displaystyle\quad+(BD+DB)xz+(CD+DC)yz+x^{2}B^{2}+y^{2}C^{2}+z^{2}D^{2}.$
(A.5)

Hence, the $xyz$-coefficient in $P^{4}(x,y,z)$ is the $xyz$-coefficient in

$\displaystyle\Big{[}(AB+BA)x+(AC+CA)y+(AD+DA)z$
$\displaystyle+(BC+CB)xy+(BD+DB)xz+(CD+DC)yz\Big{]}$ $\displaystyle\times$
$\displaystyle\Big{[}(AB+BA)x+(AC+CA)y+(AD+DA)z$
$\displaystyle+(BC+CB)xy+(BD+DB)xz+(CD+DC)yz\Big{]}.$

from which (A.1) is obtained. Hence the $xyz$-coefficient in
$tr\big{[}P^{4}(x,y,z)\big{]}$ is given by

$\displaystyle 2T$ $\displaystyle
r\Big{[}(AB+BA)(CD+DC)+(AC+CA)(BD+DB)+(AD+DA)(BC+CB)\Big{]}$
$\displaystyle=2T$ $\displaystyle
r\Big{[}ABCD+ABDC+BACD+BADC+ACBD+ACDB+CABD+CADB$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ADBC+ADCB+DABC+DACB\Big{]}$
$\displaystyle=4T$ $\displaystyle
r\Big{[}ABCD+ABDC+ACBD+ACDB+ADBC+ADCB\Big{]}.$

∎

######  Proposition A.2.

The coefficient of $xyz$ in
$tr\big{[}P^{3}(x,y,z)\big{]}.tr\big{[}P(x,y,z)\big{]}$ is given by

$3\Bigg{\\{}tr(A)tr\Big{[}BCD+BDC\Big{]}+tr(B)tr\Big{[}ACD+ADC\Big{]}+tr(C)tr\Big{[}ABD+ADB\Big{]}+tr(D)tr\Big{[}ABC+ACB\Big{]}\Bigg{\\}}.$
(A.6)

###### Proof.

By (A), we know that

$\displaystyle P^{3}(x,y,z)=\Big{[}A^{2}$
$\displaystyle+(AB+BA)x+(AC+CA)y+(AD+DA)z+(BC+CB)xy+(BD+DB)xz$
$\displaystyle+(CD+DC)yz+x^{2}B^{2}+y^{2}C^{2}+z^{2}D^{2}\Big{]}\times\Big{[}A+xB+yC+zD\Big{]}.$
(A.7)

This shows that in $P^{3}(x,y,z)$ the coefficient of the

  1. (1)

$xyz$-term is given by $(BC+CB)D+(BD+DB)C+(CD+DC)B$ and whose trace is given
by

$tr\Big{[}BCD+CBD+BDC+DBC+CDB+DCB\Big{]}=3tr\Big{[}BCD+BDC\Big{]}$ (A.8)

  2. (2)

$xy$-term is given by $(AB+BA)C+(AC+CA)B+(BC+CB)A$ and whose trace is given by

$tr\Big{[}ABC+BAC+ACB+CAB+BCA+CBA\Big{]}=3tr\Big{[}ABC+ACB\Big{]}$ (A.9)

  3. (3)

$yz$-term is given by $(AC+CA)D+(AD+DA)C+(CD+DC)A$ and whose trace is given by

$tr\Big{[}ACD+CAD+ADC+DAC+CDA+DCA\Big{]}=3tr\Big{[}ACD+ADC\Big{]}$ (A.10)

  4. (4)

$xz$-term is given by $(AB+BA)D+(AD+DA)B+(BD+DB)A$ and whose trace is given by

$tr\Big{[}ABD+BAD+ADB+DAB+BDA+DBA\Big{]}=3tr\Big{[}ABD+ADB\Big{]}.$ (A.11)

To get the coefficient of $xyz$ in
$tr\big{[}P^{3}(x,y,z)\big{]}.tr\big{[}P(x,y,z)\big{]}$, we multiply (A.8) by
$tr(A)$, (A.9) by $tr(D)$, (A.10) by $tr(B)$ and (A.11) by $tr(C)$ from which
the proposition then follows. ∎

######  Proposition A.3.

The coefficient of $xyz$ in $\bigg{[}tr\big{[}P^{2}(x,y,z)\big{]}\bigg{]}^{2}$
is given by

$8\Big{\\{}tr(AB)tr(CD)+tr(AC)tr(BD)+tr(AD)tr(BC)\Big{\\}}$ (A.12)

and the $xyz$-coefficient in
$tr\big{[}P^{2}(x,y,z)\big{]}\cdot\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{2}$
is given by

$\displaystyle 4\Big{\\{}tr(AB)tr(C)tr(D)$
$\displaystyle+tr(AC)tr(B)tr(D)+tr(AD)tr(B)tr(C)$
$\displaystyle+tr(CD)tr(A)tr(B)+tr(BD)tr(A)tr(C)+tr(BC)tr(A)tr(D)\Big{\\}}.$
(A.13)

###### Proof.

Following (A) , we have

$\displaystyle tr\big{[}P^{2}(x,y,z)\big{]}=2\big{[}$ $\displaystyle
xtr(AB)+ytr(AC)+ztr(AD)+xytr(BC)+xztr(BD)+yztr(CD)\Big{]}$ (A.14)
$\displaystyle+tr(A^{2}+x^{2}B^{2}+y^{2}C^{2}+z^{2}D^{2}).$

Squaring the preceding equality, the $xyz$ coefficient is equal to

$\displaystyle
4\big{[}tr(AB)tr(CD)+tr(AC)tr(BD)+tr(AD)tr(BC)+tr(BC)tr(AD)+tr(B$
$\displaystyle D)tr(AC)$ $\displaystyle+tr(CD)tr(AB)\big{]}.$

for which (A.12) follows. Moreover,
$\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{2}$ is given by

$\displaystyle 2\Big{[}xtr(A)tr(B)+ytr(A)tr(C)+ztr(A)tr(D)$
$\displaystyle+xytr(B)tr(C)+xztr(B)tr(D)+yztr(C)tr(D)\Big{]}$
$\displaystyle+(trA)^{2}+x^{2}(trB)^{2}+y^{2}(trC)^{2}+z^{2}(trD)^{2}.$

Multiplying the above equation by (A.14) we get (A.3). ∎

Finally, we note that

$\bigg{[}tr\big{[}P(x,y,z)\big{]}\bigg{]}^{4}=\bigg{[}tr(A)+xtr(B)+yr(C)+ztr(D)\bigg{]}^{4}=\sum_{\begin{subarray}{c}|\alpha|=4\\\
\alpha\in\mathbb{N}_{0}^{4}\end{subarray}}\binom{4}{\alpha}\Big{(}tr(A),xtr(B),ytr(C),ztr(D)\Big{)}^{\alpha}.$

So that for $\alpha=(1,1,1,1)$ we obtain the coefficient of the
$xyz$-variable:

$24tr(A)tr(B)tr(C)tr(D).$ (A.15)

Multiplying (A.4) by 6, (A.6) by 8, (A.12) by 3, (A.3) by 6 and then adding
all together to (A.15) we get (A.2).