$|g_{t}^{(\star,\varepsilon)}-g_{t}^{\star}|=|{\langle{{\mathrm{g}}^{(\star,\varepsilon)}-{\mathrm{g}}^{\star},\mathbbm{k}_{t}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}|\leq\|{\mathrm{g}}^{(\star,\varepsilon)}-{\mathrm{g}}^{\star}\|_{{\mathcal{H}}_{\mathbbm{k}}}\|\mathbbm{k}_{t}\|_{{\mathcal{H}}_{\mathbbm{k}}}.$

Accordingly, from $\|\mathbbm{k}_{t}\|=\mathbbm{k}(t,t)^{\frac{1}{2}}$, it
follows that

$|g_{t}^{(\star,\varepsilon)}-g_{t}^{\star}|\leq\|{\mathrm{g}}^{(\star,\varepsilon)}-{\mathrm{g}}^{\star}\|_{{\mathcal{H}}_{\mathbbm{k}}}\sup_{t\in{\mathbb{T}}}\mathbbm{k}(t,t)^{\frac{1}{2}}.$

Since
$\sup_{t\in{\mathbb{T}}}\mathbbm{k}(t,t)^{\frac{1}{2}}\\!\\!<\\!\\!\infty$ and
$\lim_{\varepsilon\to\mathbf{0}}\|{\mathrm{g}}^{(\star,\varepsilon)}-{\mathrm{g}}^{\star}\|_{{\mathcal{H}}_{\mathbbm{k}}}=0$,
we have
$g_{t}^{(\star,\varepsilon)}\operatornamewithlimits{\longrightarrow}\limits^{\varepsilon\to
0}g_{t}^{\star}$, uniformly in ${\mathbb{T}}$, and proof concludes. ∎

###  A-G Optimization Problem for TC Kernel

Let ${\mathbb{T}}={\mathbb{R}}_{+}$ and
$\mathbbm{k}(s,t):=\mathrm{e}^{-\beta\max(s,t)}$, for
$s,t\in{\mathbb{R}}_{+}$, where $\beta$ is a positive real scalar.
Accordingly, we define $\varphi_{\omega,t}$ as
$\varphi_{\omega,t}:=\int_{{\mathbb{R}}_{+}}\mathbbm{k}(t,s)\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s$, for $\omega,t\in{\mathbb{R}}_{+}$.

######  Lemma 16.

For any $\omega,t\in{\mathbb{R}}_{+}$, we have

$\varphi_{\omega,t}=\begin{cases}\mathrm{e}^{-\beta
t}\Big{[}\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-\mathrm{j}\omega
t}}{\beta+\mathrm{j}\omega}\Big{]},&\text{ if }\omega\neq 0,\\\
\mathrm{e}^{-\beta t}(t+\frac{1}{\beta}),&\text{ if }\omega=0.\\\ \end{cases}$
(82)

###### Proof.

When $\omega\neq 0$, we have

$\begin{split}&\varphi_{\omega,t}=\int_{0}^{\infty}\\!\mathbbm{k}(t,s)\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s\\\
&\\!=\int_{0}^{t}\\!\mathrm{e}^{-\beta\max(t,s)}\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s+\int_{t}^{\infty}\\!\mathrm{e}^{-\beta\max(t,s)}\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s\\\ &\\!=\int_{0}^{t}\\!\mathrm{e}^{-\beta
t}\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s+\int_{t}^{\infty}\\!\mathrm{e}^{-\beta
s}\mathrm{e}^{-\mathrm{j}\omega s}\mathrm{d}s\\\ &\\!=\mathrm{e}^{-\beta
t}\int_{0}^{t}\\!\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s+\int_{t}^{\infty}\\!\mathrm{e}^{-(\beta+\mathrm{j}\omega)s}\mathrm{d}s\\\
&\\!=\mathrm{e}^{-\beta t}\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-(\beta+\mathrm{j}\omega)t}}{\beta+\mathrm{j}\omega}\\\
&\\!=\mathrm{e}^{-\beta t}\bigg{[}\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-\mathrm{j}\omega
t}}{\beta+\mathrm{j}\omega}\bigg{]}.\end{split}$

The case $\omega=0$ follows from similar line of arguments. ∎

######  Theorem 17.

For any $\omega,t\in{\mathbb{R}}_{+}$, we have

$\begin{split}&\varphi_{\omega,t}^{\text{\rm{(r)}}}=\begin{cases}\mathrm{e}^{-\beta
t}\ \mathrm{real}(\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-\mathrm{j}\omega
t}}{\beta+\mathrm{j}\omega}),&\text{ if }\omega\neq 0,\\\ \mathrm{e}^{-\beta
t}(t+\frac{1}{\beta}),&\text{ if }\omega=0,\\\ \end{cases}\\\
&\varphi_{\omega,t}^{\text{\rm{(i)}}}=\begin{cases}\mathrm{e}^{-\beta t}\
\mathrm{imag}(\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-\mathrm{j}\omega
t}}{\beta+\mathrm{j}\omega}),&\text{ if }\omega\neq 0,\\\ 0,&\text{ if
}\omega=0.\\\ \end{cases}\end{split}$ (83)

###### Proof.

From Lemma 3 and definition of $\varphi_{\omega,t}^{\text{\rm{(r)}}}$, it
follows that

$\begin{split}\varphi_{\omega,t}^{\text{\rm{(r)}}}&={\langle{\varphi_{\omega}^{\text{\rm{(r)}}},\mathbbm{k}_{t}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}\\\
&=\int_{{\mathbb{R}}_{+}}\\!\mathbbm{k}(t,s)\cos(\omega s)\mathrm{d}s\\\
&=\mathrm{real}(\int_{{\mathbb{R}}_{+}}\\!\mathbbm{k}(t,s)\mathrm{e}^{-\mathrm{j}\omega
s}\mathrm{d}s)=\mathrm{real}(\varphi_{\omega,t}).\end{split}$

Similarly, we have
$\varphi_{\omega,t}^{\text{\rm{(i)}}}=\mathrm{imag}(\varphi_{\omega,t})$.
Following these, the claim concluded form Lemma 16. ∎

For any $\omega_{1},\omega_{2}\in{\mathbb{R}}_{+}$, define
$\zeta_{r}(\omega_{1},\omega_{2})$ and $\zeta_{i}(\omega_{1},\omega_{2})$ as

$\begin{split}\zeta_{\text{r}}(\omega_{1},\omega_{2})&:=\int_{{\mathbb{R}}_{+}}\varphi_{\omega_{2},t}^{\text{\rm{(r)}}}\mathrm{e}^{-\mathrm{j}\omega_{1}t}\mathrm{d}t,\\\
\zeta_{\text{i}}(\omega_{1},\omega_{2})&:=\int_{{\mathbb{R}}_{+}}\varphi_{\omega_{2},t}^{\text{\rm{(i)}}}\mathrm{e}^{-\mathrm{j}\omega_{1}t}\mathrm{d}t.\end{split}$

######  Lemma 18.

Let $\omega_{1},\omega_{2}\in{\mathbb{R}}_{+}$. If $\omega_{2}\neq 0$, then we
have

$\begin{split}&\zeta_{\text{r}}(\omega_{1},\omega_{2})=\frac{\beta}{2}\bigg{[}\frac{1}{(\omega_{2}^{2}-\mathrm{j}\omega_{2}\beta)(\beta+\mathrm{j}\omega_{1}+\mathrm{j}\omega_{2})}\\\
&\quad+\frac{1}{(\omega_{2}^{2}+\mathrm{j}\omega_{2}\beta)(\beta+\mathrm{j}\omega_{1}-\mathrm{j}\omega_{2})}\bigg{]},\\\
\end{split}$ (84)

and

$\begin{split}&\zeta_{\text{i}}(\omega_{1},\omega_{2})=\frac{\beta}{2\mathrm{j}}\bigg{[}\frac{1}{(\omega_{2}^{2}-\mathrm{j}\omega_{2}\beta)(\beta+\mathrm{j}\omega_{1}+\mathrm{j}\omega_{2})}\\\
&\quad-\frac{1}{(\omega_{2}^{2}+\mathrm{j}\omega_{2}\beta)(\beta+\mathrm{j}\omega_{1}-\mathrm{j}\omega_{2})}\bigg{]}-\frac{1}{\omega_{2}}\frac{1}{(\beta+\mathrm{j}\omega_{1})}.\end{split}$
(85)

If $\omega_{2}=0$, then $\zeta_{\text{i}}(\omega_{1},\omega_{2})=0$ and
$\zeta_{\text{r}}(\omega_{1},\omega_{2})=\frac{2\beta+\mathrm{j}\omega_{1}}{\beta(\beta+\mathrm{j}\omega_{1})^{2}}.$

###### Proof.

Note that for any $z\in{\mathbb{C}}$, we have
$\mathrm{real}(z)=\frac{1}{2}(z+z^{*})$, where $z^{*}$ denotes the complex
conjugate of $z$. Accordingly, when $\omega_{2}\neq 0$, from (83) and the
definition of $\zeta_{\text{r}}$, we have

$\begin{split}&\\!\\!\zeta_{\text{r}}(\omega_{1},\omega_{2})\\!=\\!\frac{1}{2}\\!\int_{{\mathbb{R}}_{+}}\\!\\!\\!\Bigg{[}\\!\bigg{[}\frac{\mathrm{e}^{-\beta
t}}{\mathrm{j}\omega_{2}}\\!+\\!\big{(}\frac{1}{\beta+\mathrm{j}\omega_{2}}\\!-\\!\frac{1}{\mathrm{j}\omega_{2}}\big{)}\mathrm{e}^{-(\beta+\mathrm{j}\omega_{2})t}\bigg{]}\\\
&\quad+\bigg{[}\\!-\\!\frac{\mathrm{e}^{-\beta
t}}{\mathrm{j}\omega_{2}}\\!+\\!\big{(}\frac{1}{\beta-\mathrm{j}\omega_{2}}\\!+\\!\frac{1}{\mathrm{j}\omega_{2}}\big{)}\mathrm{e}^{-(\beta-\mathrm{j}\omega_{2})t}\bigg{]}\\!\Bigg{]}\mathrm{e}^{-\mathrm{j}\omega_{1}t}\mathrm{d}t\\\
&=\frac{1}{2}\int_{{\mathbb{R}}_{+}}\\!\bigg{[}\big{(}\frac{1}{\beta+\mathrm{j}\omega_{2}}-\frac{1}{\mathrm{j}\omega_{2}}\big{)}\mathrm{e}^{-(\beta+\mathrm{j}\omega_{1}+\mathrm{j}\omega_{2})t}\\\
&\quad+\big{(}\frac{1}{\beta-\mathrm{j}\omega_{2}}+\frac{1}{\mathrm{j}\omega_{2}}\big{)}\mathrm{e}^{-(\beta+\mathrm{j}\omega_{1}-\mathrm{j}\omega_{2})t}\bigg{]}\mathrm{d}t.\end{split}$

Accordingly, we have

$\begin{split}\zeta_{\text{r}}(\omega_{1},\omega_{2})=&\frac{1}{2}\big{(}\frac{1}{\beta+\mathrm{j}\omega_{2}}-\frac{1}{\mathrm{j}\omega_{2}}\big{)}\frac{1}{\beta+\mathrm{j}\omega_{1}+\mathrm{j}\omega_{2}}\\\
&\quad+\frac{1}{2}\big{(}\frac{1}{\beta-\mathrm{j}\omega_{2}}+\frac{1}{\mathrm{j}\omega_{2}}\big{)}\frac{1}{\beta+\mathrm{j}\omega_{1}-\mathrm{j}\omega_{2}}.\end{split}$

Subsequently, we have (84). The claim for the other cases can be proved based
on similar line of arguments. ∎

######  Theorem 19.

For any $\omega_{1},\omega_{2}\in{\mathbb{R}}_{+}$, we have

$\begin{split}&{\langle{\varphi_{\omega_{1}}^{\text{\rm{(r)}}}\\!,\\!\varphi_{\omega_{2}}^{\text{\rm{(r)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\mathrm{real}(\zeta_{\text{r}}(\omega_{1}\\!,\omega_{2}))_{{\mathcal{H}}_{\mathbbm{k}}},\\\
&{\langle{\varphi_{\omega_{1}}^{\text{\rm{(r)}}}\\!,\\!\varphi_{\omega_{2}}^{\text{\rm{(i)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\mathrm{real}(\zeta_{\text{i}}(\omega_{1}\\!,\omega_{2}))_{{\mathcal{H}}_{\mathbbm{k}}},\\\
&{\langle{\varphi_{\omega_{1}}^{\text{\rm{(i)}}}\\!,\\!\varphi_{\omega_{2}}^{\text{\rm{(r)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\mathrm{imag}(\zeta_{\text{r}}(\omega_{1}\\!,\omega_{2}))_{{\mathcal{H}}_{\mathbbm{k}}},\\\
&{\langle{\varphi_{\omega_{1}}^{\text{\rm{(i)}}}\\!,\\!\varphi_{\omega_{2}}^{\text{\rm{(i)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\mathrm{imag}(\zeta_{\text{i}}(\omega_{1}\\!,\omega_{2}))_{{\mathcal{H}}_{\mathbbm{k}}}.\end{split}$

###### Proof.

This is a straightforward result from (18), (15), and the definition of
$\zeta_{r}$ and $\zeta_{i}$. ∎

######  Theorem 20.

Define $z_{{\mathrm{u}}}(\omega,\tau)$ as

$\begin{split}\\!\\!\\!\\!\\!\\!&z_{{\mathrm{u}}}(\omega,\tau)\\!:=\\!\begin{cases}\int_{{\mathbb{R}}_{+}}\\!\\!\big{(}\frac{\mathrm{e}^{-\beta
t}}{\mathrm{j}\omega}+\frac{\beta\mathrm{e}^{-(\beta+\mathrm{j}\omega)t}}{\omega^{2}-\mathrm{j}\omega\beta}\big{)}u_{\tau-t}\mathrm{d}t,\\!\\!\\!\\!\\!\\!&\text{
if }\omega\\!\neq\\!0,\\\ \int_{{\mathbb{R}}_{+}}\\!\\!\mathrm{e}^{-\beta
t}(t+\frac{1}{\beta})u_{\tau-t}\mathrm{d}t,\\!\\!\\!\\!\\!\\!&\text{ if
}\omega\\!=\\!0.\\\ \end{cases}\end{split}$ (86)

for $\omega,\tau\in{\mathbb{R}}_{+}$. Then, we have

$\displaystyle{\langle{\varphi_{\omega}^{\text{\rm{(r)}}},\varphi_{\tau}^{\text{\rm{(u)}}}}\rangle}=\mathrm{real}(z_{{\mathrm{u}}}(\beta,\omega,\tau)),$
(87)
$\displaystyle{\langle{\varphi_{\omega}^{\text{\rm{(i)}}},\varphi_{\tau}^{\text{\rm{(u)}}}}\rangle}=\mathrm{imag}(z_{{\mathrm{u}}}(\beta,\omega,\tau)).$
(88)

###### Proof.

When $\omega\neq 0$, from (83), (2) and Lemma 5, we have

$\begin{split}&{\langle{\varphi_{\omega}^{\text{\rm{(r)}}},\varphi_{\tau}^{\text{\rm{(u)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\int_{0}^{\infty}\\!\varphi_{\omega,t}^{\text{\rm{(r)}}}u_{\tau-t}\mathrm{d}t\\\
&=\int_{0}^{\infty}\\!\mathrm{e}^{-\beta t}\
\mathrm{real}(\frac{1-\mathrm{e}^{-\mathrm{j}\omega
t}}{\mathrm{j}\omega}+\frac{\mathrm{e}^{-\mathrm{j}\omega
t}}{\beta+\mathrm{j}\omega})u_{\tau-t}\mathrm{d}t\\\
&=\mathrm{real}(\int_{0}^{\infty}\\!\big{(}\frac{1}{\mathrm{j}\omega}\mathrm{e}^{-\beta
t}+\frac{\beta}{\omega^{2}-\mathrm{j}\omega\beta}\mathrm{e}^{-(\beta+\mathrm{j}\omega)t}\big{)}u_{\tau-t}\mathrm{d}t).\end{split}$

Similarly, when $\omega=0$, we have

${\langle{\varphi_{\omega}^{\text{\rm{(r)}}},\varphi_{\tau}^{\text{\rm{(u)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}\\!=\\!\int_{{\mathbb{R}}_{+}}\\!\\!\varphi_{\omega,t}^{\text{\rm{(r)}}}u_{\tau-t}\mathrm{d}t\\!=\\!\int_{{\mathbb{R}}_{+}}\\!\\!\mathrm{e}^{-\beta
t}(t+\frac{1}{\beta})u_{\tau-t}\mathrm{d}t.$

The other equality can be shown in a similar way. ∎

######  Remark 3.

When ${\mathbb{T}}={\mathbb{Z}}_{+}$, using similar lines of arguments, one
can introduce results analogous to Lemma 16, Proposition 17, Lemma 18,
Proposition 19 and Proposition 20.

######  Remark 4.

For the case of ${\mathbb{T}}={\mathbb{Z}}_{+}$, for
$z_{{\mathrm{u}}}(\omega,\tau)$, we have a finite summation in a similar form
to (86), when the system is initially at rest, that is $u_{t}=0$, for $t<0$.

The set of piecewise constant functions are dense in $L^{p}$, i.e., one can
approximate precisely almost any signal of interest using a piecewise constant
signal. Motivated by this fact, let assume the input signal
${\mathrm{u}}=(u_{t})_{t\in{\mathbb{R}}_{+}}$ is given as a piecewise constant
function defined as

$u_{t}=\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\mathbf{1}_{[s_{i},s_{i+1})}(t),\quad\forall
t\in{\mathbb{R}}_{+},$ (89)

where $n_{\text{s}}\in{\mathbb{N}}$, $\xi_{i}\in{\mathbb{R}}$, for
$i=0,\ldots,n_{\text{s}}$, and $(s_{0},s_{1},\ldots,s_{n_{\text{s}}})$ is a
finite increasing sequence in ${\mathbb{R}}_{+}$ such that $s_{0}=0$.

######  Theorem 21.

Let $\tau\in{\mathbb{R}}_{+}$. Then, if $\omega\neq 0$, we have

$\begin{split}z_{{\mathrm{u}}}&(\omega,\tau)=\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\bigg{[}\frac{\mathrm{e}^{-\beta\bar{s}_{i+1}(\tau)}-\mathrm{e}^{-\beta\bar{s}_{i}(\tau)}}{\mathrm{j}\omega\beta}\\\
&\quad+\frac{\beta\mathrm{e}^{-(\beta+\mathrm{j}\omega)\bar{s}_{i+1}(\tau)}-\beta\mathrm{e}^{-(\beta+\mathrm{j}\omega)\bar{s}_{i}(\tau)}}{(\omega^{2}-\mathrm{j}\omega\beta)(\beta+\mathrm{j}\omega)}\bigg{]},\end{split}$
(90)

and, if $\omega=0$, we have

$\begin{split}&z_{{\mathrm{u}}}(\omega,\tau)=\sum_{i=0}^{n_{\text{s}}-1}\frac{\xi_{i+1}}{\beta^{2}}\bigg{[}(\beta\bar{s}_{i+1}(\tau)\\!+\\!2)\mathrm{e}^{-\beta\bar{s}_{i+1}(\tau)}\\\
&\qquad\qquad\qquad\qquad\qquad-(\beta\bar{s}_{i}(\tau)\\!+\\!2)\mathrm{e}^{-\beta\bar{s}_{i}(\tau)}\bigg{]},\end{split}$
(91)

where $\bar{s}_{i}(\tau):=\max(\tau-s_{i},0)$, for $i=0,\ldots,n_{\text{s}}$.

###### Proof.

From (89) and (86), if $\omega\neq 0$, we have

$\begin{split}z_{{\mathrm{u}}}&(\omega,\tau)\\\
&=\int_{{\mathbb{R}}_{+}}\left[\frac{\mathrm{e}^{-\beta
t}}{\mathrm{j}\omega}+\frac{\beta\mathrm{e}^{-(\beta+\mathrm{j}\omega)t}}{\omega^{2}-\mathrm{j}\omega\beta}\right]\sum_{i=0}^{n_{\text{s}}-1}\\!\xi_{i+1}\mathbf{1}_{[s_{i},s_{i+1})}(\tau-t)\mathrm{d}t\\\
&=\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\int_{\bar{s}_{i+1}(\tau)}^{\bar{s}_{i}(\tau)}\frac{\mathrm{e}^{-\beta
t}}{\mathrm{j}\omega}\mathrm{d}t+\int_{\bar{s}_{i+1}(\tau)}^{\bar{s}_{i}(\tau)}\frac{\beta\mathrm{e}^{-(\beta+\mathrm{j}\omega)t}}{\omega^{2}-\mathrm{j}\omega\beta}\mathrm{d}t,\end{split}$

which concludes (90) when we replace the integrals with their closed-form
values. Similarly, one can show (91). ∎

Note that using (90) and (91), the inner products in Proposition 20 can be
obtained as a finite sum rather than an improper integral.

Define
$\psi:{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\to{\mathbb{R}}$
as

$\begin{split}&\psi(t,a,b):=\frac{1}{\beta}(\mathrm{e}^{-\beta\max(\min(t,b),a)}-\mathrm{e}^{-\beta
b})\\\
&\qquad\qquad\qquad+\bigg{[}\max(\min(t,b),a)-a\bigg{]}\mathrm{e}^{-\beta
t}.\end{split}$ (92)

for any $a,b,t\in{\mathbb{R}}_{+}$.

######  Lemma 22.

For any $a,b\in{\mathbb{R}}_{+}$ such that $a\leq b$, we have

$\psi(t,a,b)=\int_{a}^{b}\\!\mathrm{e}^{-\beta\max(t,s)}\mathrm{d}s.$ (93)

###### Proof.

Considering three cases of $t<a$, $t>b$ and $t\in[a,b]$ and then evaluating
the integral $\int_{a}^{b}\\!\mathrm{e}^{-\beta\max(t,s)}\mathrm{d}s$, one can
obtain (92) by comparing the resulting values. ∎

######  Theorem 23.

For any $t\in{\mathbb{R}}_{+}$, we have

$\varphi_{\tau,t}^{\text{\rm{(u)}}}=\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\psi(t,\bar{s}_{i+1}(\tau),\bar{s}_{i}(\tau)).$
(94)

###### Proof.

From (89) and Lemma 5, we have

$\begin{split}\varphi_{\tau,t}^{\text{\rm{(u)}}}&={\langle{\varphi_{\tau}^{\text{\rm{(u)}}},\mathbbm{k}_{t}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}\\\
&=\int_{{\mathbb{R}}_{+}}\mathbbm{k}(t,s)u(\tau-s)\mathrm{d}s\\\
&=\int_{{\mathbb{R}}_{+}}\mathbbm{k}(t,s)\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\mathbf{1}_{[s_{i},s_{i+1})}(\tau-s)\mathrm{d}s\\\
&=\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\int_{\bar{s}_{i+1}(\tau)}^{\bar{s}_{i}(\tau)}\\!\mathrm{e}^{-\beta\max(t,s)}\mathrm{d}s.\end{split}$
(95)

Subsequently, the claim follows from Lemma 22. ∎

######  Lemma 24.

Define function $\nu:{\mathbb{R}}_{\geq 0}\times{\mathbb{R}}_{\geq
0}\to{\mathbb{R}}$ as

$\nu(x,y):=\int_{0}^{x}\\!\int_{0}^{y}\\!\mathbbm{k}(s,t)\mathrm{d}t\
\\!\mathrm{d}s.$ (96)

Then, we have

$\begin{split}\nu(x,y)=\frac{1}{\beta^{2}}\Big{[}2&-2\mathrm{e}^{-\beta\min(x,y)}\\\
&\qquad-\beta\min(x,y)(\mathrm{e}^{-\beta x}+\mathrm{e}^{-\beta
y})\Big{]}.\end{split}$ (97)

###### Proof.

Since $\mathbbm{k}(s,t)=\mathbbm{k}(t,s)$, without loss of generality, let
assume $x\leq y$. Then, we have

$\begin{split}\nu&(x,y)=\int_{0}^{x}\\!\int_{0}^{y}\\!\mathbbm{k}(s,t)\mathrm{d}t\
\\!\mathrm{d}s\\\
&=\int_{0}^{x}\\!\int_{0}^{y}\\!\mathrm{e}^{-\beta\max(s,t)}\mathrm{d}t\
\\!\mathrm{d}s\\\
&=\int_{0}^{x}\\!\int_{0}^{s}\\!\mathrm{e}^{-\beta\max(s,t)}\mathrm{d}t\
\\!\mathrm{d}s+\int_{0}^{x}\\!\int_{s}^{y}\\!\mathrm{e}^{-\beta\max(s,t)}\mathrm{d}t\
\\!\mathrm{d}s\\\ &=\int_{0}^{x}\\!\int_{0}^{s}\\!\mathrm{e}^{-\beta
s}\mathrm{d}t\ \\!\mathrm{d}s+\int_{0}^{x}\\!\int_{s}^{y}\\!\mathrm{e}^{-\beta
t}\mathrm{d}t\ \\!\mathrm{d}s\\\ &=\int_{0}^{x}\\!s\mathrm{e}^{-\beta
s}\mathrm{d}s+\int_{0}^{x}\\!\frac{1}{\beta}(\mathrm{e}^{-\beta
s}-\mathrm{e}^{-\beta y})\mathrm{d}s\\\ &=\int_{0}^{x}\\!s\mathrm{e}^{-\beta
s}\mathrm{d}s+\frac{1}{\beta^{2}}(1-\mathrm{e}^{-\beta
x})-\frac{x}{\beta}\mathrm{e}^{-\beta y}\int_{0}^{x}\\!s\mathrm{e}^{-\beta
s}\mathrm{d}s\\\
&=\frac{1}{\beta^{2}}-\frac{1}{\beta}(x+\frac{1}{\beta})\mathrm{e}^{-\beta
x}+\frac{1}{\beta^{2}}(1-\mathrm{e}^{-\beta
x})-\frac{x}{\beta}\mathrm{e}^{-\beta y}\\\
&=\frac{1}{\beta^{2}}\bigg{(}2-2\mathrm{e}^{-\beta
x}-x\beta(\mathrm{e}^{-\beta x}+\mathrm{e}^{-\beta y})\bigg{)}.\end{split}$

Accordingly, since $x=\min(x,y)$ and $y=\max(x,y)$, one has (97). ∎

For each $i,j\in\\{0,1,\ldots,n_{\text{s}}-1\\}$, define function
$\kappa_{ij}:{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\to{\mathbb{R}}_{\geq 0}$
such that, for any $\tau_{1},\tau_{2}\in{\mathbb{R}}_{+}$, we have

$\begin{split}&\kappa_{ij}(\tau_{1},\tau_{2})=\nu(\bar{s}_{i}(\tau_{1}),\bar{s}_{j}(\tau_{2}))-\nu(\bar{s}_{i+1}(\tau_{1}),\bar{s}_{j}(\tau_{2}))\\\
&\quad-\nu(\bar{s}_{i}(\tau_{1}),\bar{s}_{j+1}(\tau_{2}))+\nu(\bar{s}_{i+1}(\tau_{1}),\bar{s}_{j+1}(\tau_{2})).\end{split}$
(98)

######  Theorem 25.

For any $\tau_{1},\tau_{2}\in{\mathbb{R}}_{+}$, we have

$\begin{split}&{\langle{\varphi_{\tau_{1}}^{\text{\rm{(u)}}},\varphi_{\tau_{2}}^{\text{\rm{(u)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}=\sum_{i=0}^{n_{\text{s}}-1}\sum_{j=0}^{n_{\text{s}}-1}\xi_{i+1}\xi_{j+1}\kappa_{ij}(\tau_{1},\tau_{2}).\end{split}$
(99)

###### Proof.

Due to (96), we have

$\begin{split}&\\!\\!\\!\int_{0}^{\infty}\\!\\!\\!\int_{0}^{\infty}\\!\\!\mathbbm{k}(s,t)\mathbf{1}_{[s_{j},s_{j+1})}(\tau_{2}-t)\mathbf{1}_{[s_{i},s_{i+1})}(\tau_{1}-s)\mathrm{d}t\
\\!\mathrm{d}s\\\
&=\int_{\bar{s}_{i+1}(\tau_{1})}^{\bar{s}_{i}(\tau_{1})}\\!\int_{\bar{s}_{j+1}(\tau_{2})}^{\bar{s}_{j}(\tau_{2})}\\!\mathbbm{k}(s,t)\mathrm{d}t\
\\!\mathrm{d}s\\\
&=\nu(\bar{s}_{i}(\tau_{1}),\bar{s}_{j}(\tau_{2}))-\nu(\bar{s}_{i+1}(\tau_{1}),\bar{s}_{j}(\tau_{2}))\\\
&\quad-\nu(\bar{s}_{i}(\tau_{1}),\bar{s}_{j+1}(\tau_{2}))+\nu(\bar{s}_{i+1}(\tau_{1}),\bar{s}_{j+1}(\tau_{2})).\end{split}$
(100)

From other hand, due to Lemma 5 and (89), we have

$\begin{split}&{\langle{\varphi_{\tau_{1}}^{\text{\rm{(u)}}},\varphi_{\tau_{2}}^{\text{\rm{(u)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}\\\
&=\int_{0}^{\infty}\varphi_{\tau_{2},s}^{\text{\rm{(u)}}}\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\mathbf{1}_{[s_{i},s_{i+1})}(\tau_{1}-s)\mathrm{d}s\\\
&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbbm{k}(s,t)\sum_{j=0}^{n_{\text{s}}-1}\xi_{j+1}\mathbf{1}_{[s_{j},s_{j+1})}(\tau_{2}-t)\\\
&\qquad\qquad\sum_{i=0}^{n_{\text{s}}-1}\xi_{i+1}\mathbf{1}_{[s_{i},s_{i+1})}(\tau_{1}-s)\mathrm{d}t\
\\!\mathrm{d}s\\\
&=\sum_{i=0}^{n_{\text{s}}-1}\sum_{j=0}^{n_{\text{s}}-1}\xi_{i+1}\xi_{j+1}\bigg{[}\int_{0}^{\infty}\int_{0}^{\infty}\mathbbm{k}(s,t)\\\
&\qquad\qquad\mathbf{1}_{[s_{j},s_{j+1})}(\tau_{2}-t)\mathbf{1}_{[s_{i},s_{i+1})}(\tau_{1}-s)\mathrm{d}t\
\\!\mathrm{d}s\bigg{]}.\end{split}$ (101)

Accordingly, one can conclude (99) due to (100). ∎

######  Theorem 26.

Let ${\mathbb{T}}={\mathbb{Z}}_{+}$ and the system be initially at rest, i.e.,
we have $u_{t}=0$, for $t<0$. Also, let assume that $t_{i}=i$, for
$i=0,\ldots,n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}-1$. Define matrix
${\mathrm{K}}\in{\mathbb{R}}^{n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}\times
n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}}$ as
${\mathrm{K}}:=\begin{bmatrix}\mathbbm{k}(t-1,s-1)\end{bmatrix}_{t=1,s=1}^{n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}},n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}}$
and Toeplitz matrix
${\mathrm{T}}_{{\mathrm{u}}}\in{\mathbb{R}}^{n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}\times
n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}}$ as
${\mathrm{T}}_{{\mathrm{u}}}:=\begin{bmatrix}u_{t-s}\end{bmatrix}_{t=1,s=1}^{n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}},n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}}$.
Then, we have

$\Big{[}{\langle{\varphi_{i-1}^{\text{\rm{(u)}}},\varphi_{j-1}^{\text{\rm{(u)}}}}\rangle}_{{\mathcal{H}}_{\mathbbm{k}}}\Big{]}_{i=1,j=1}^{n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}},n_{{\scalebox{0.5}{$\\!{\mathscr{D}}$}}}}={\mathrm{T}}_{{\mathrm{u}}}{\mathrm{K}}{\mathrm{T}}_{{\mathrm{u}}}^{\mathsf{\scriptscriptstyle
T}}.$ (102)

###### Proof.

The claim is a straightforward result of (23) and the definition of matrices
${\mathrm{K}}$ and ${\mathrm{T}}_{{\mathrm{u}}}$. ∎

###  A-H Utilizing Barrier Method

Let $p:{\mathbb{R}}^{m}\to{\mathbb{R}}$ and
$p_{j}:{\mathbb{R}}^{m}\to{\mathbb{R}}$, for
$j\in{\mathcal{I}}^{{\scalebox{0.63}{$(\mathscr{P})$}}}$, be respectively
defined as

$p({\mathrm{x}}):=\frac{1}{2}\|{\mathrm{A}}{\mathrm{x}}-{\mathrm{y}}\|^{2}+\frac{\lambda}{2}{\mathrm{x}}^{\mathsf{\scriptscriptstyle
T}}\Phi{\mathrm{x}},$ (103)

and

$p_{j}({\mathrm{x}}):=\frac{1}{2}{\mathrm{x}}^{\mathsf{\scriptscriptstyle
T}}({\mathrm{b}}_{j}{\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}+{\mathrm{c}}_{j}{\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}){\mathrm{x}}-\frac{1}{2}(1-\varepsilon),$ (104)

for all ${\mathrm{x}}\in{\mathbb{R}}^{m}$. Then, the optimization problem (56)
can be written as

$\operatornamewithlimits{min}\limits_{{\mathrm{x}}\in{\mathbb{R}}^{m}}\
p({\mathrm{x}})+\sum_{j\in{\mathcal{I}}^{{\scalebox{0.63}{$(\mathscr{P})$}}}}{\mathcal{I}}(p_{j}({\mathrm{x}})),$
(105)

where ${\mathcal{I}}:{\mathbb{R}}\to{\mathbb{R}}_{+}\cup\\{+\infty\\}$ is
defined as ${\mathcal{I}}(x):=\delta_{(-\infty,0]}(x)$, for all
$x\in{\mathbb{R}}$. The solution approach utilizing logarithmic barrier
function approximates ${\mathcal{I}}$ in (105) successively and solves the
resulting convex program to obtain a sequence in ${\mathbb{R}}^{m}$ converging
to the solution of (105). More precisely, for $n\in{\mathbb{N}}$, let
$b_{n}:{\mathbb{R}}\to{\mathbb{R}}_{+}\cup\\{+\infty\\}$ be the function
defined as $b_{n}(x):=-\theta_{n}\ln(-x)$, for $x\in{\mathbb{R}}$, where
$\\{\theta_{n}\\}_{n=1}^{\infty}$ is a decreasing sequence such that
$\theta_{n}\downarrow 0$. Define function
$f_{n}:{\mathbb{R}}\to{\mathbb{R}}_{+}\cup\\{+\infty\\}$ as following

$f_{n}({\mathrm{x}}):=p({\mathrm{x}})-\theta_{n}\sum_{j\in{\mathcal{I}}^{{\scalebox{0.63}{$(\mathscr{P})$}}}}\ln(-p_{j}({\mathrm{x}})),\quad\forall{\mathrm{x}}\in{\mathbb{R}}^{m},$
(106)

for any $n\in{\mathbb{N}}$. The gradient of $f_{n}$ is

$\begin{split}&\nabla
f_{n}({\mathrm{x}})={\mathrm{A}}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{A}}{\mathrm{x}}+\lambda\Phi{\mathrm{x}}+{\mathrm{q}}\\\
&\qquad-\theta_{n}\sum_{j=0}^{n_{{\scalebox{0.5}{$\\!{\mathscr{P}}$}}}}\frac{({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{b}}_{j}+({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{c}}_{j}}{r+\frac{1}{2}({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2}+\frac{1}{2}({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2}},\end{split}$ (107)

and, the Hessian of $f_{n}$ is

$\begin{split}\\!\\!\\!\\!\\!\\!\\!\\!&\\!\\!\\!\nabla^{2}\\!f_{n}({\mathrm{x}})\\!=\\!{\mathrm{A}}^{\mathsf{\scriptscriptstyle
T}}\\!{\mathrm{A}}\\!+\\!\lambda\Phi\\!-\\!\theta_{n}\\!\sum_{j=0}^{n_{{\scalebox{0.5}{$\\!{\mathscr{P}}$}}}}\frac{{\mathrm{b}}_{j}{\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}+{\mathrm{c}}_{j}{\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}}{r+\frac{1}{2}({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2}+\frac{1}{2}({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2}}\\\
&\\!\\!\\!+\\!\theta_{n}\\!\sum_{j=0}^{n_{{\scalebox{0.5}{$\\!{\mathscr{P}}$}}}}\frac{\left(({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{b}}_{j}+({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{c}}_{j}\right)\left(({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{b}}_{j}+({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}}){\mathrm{c}}_{j}\right)^{\mathsf{\scriptscriptstyle
T}}\\!\\!}{(r+\frac{1}{2}({\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2}+\frac{1}{2}({\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle
T}}{\mathrm{x}})^{2})^{2}}.\end{split}$ (108)

Given the gradient and the Hessian of $f_{n}$ in (107) and (108), one may
employ an iterative optimization scheme such as Newton method or (L-)BFGS
method [58] to find ${\mathrm{x}}_{n}$, the solution of the convex program
$\min_{{\mathrm{x}}\in{\mathbb{R}}^{m}}f_{n}({\mathrm{x}})$. Note that
${\mathrm{x}}=\mathbf{0}$ is feasible in (56). Also, one can see that the
feasible set in (56) is a convex set with non-empty interior, and
consequently, it is guaranteed [58] that
$\\{{\mathrm{x}}_{n}\\}_{n=1}^{\infty}$ converges to the solution of (105).

Given $\Phi{\mathrm{x}}$, we have the value of ${\mathrm{A}}{\mathrm{x}}$ as
well as ${\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle T}}{\mathrm{x}}$ and
${\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle T}}{\mathrm{x}}$, for
$j\in{\mathcal{I}}^{{\scalebox{0.63}{$(\mathscr{P})$}}}$. Accordingly, by
obtaining the vector $\Phi{\mathrm{x}}$, one can calculate the value in (107)
efficiently. Moreover, one can see from (108) that the matrices
${\mathrm{A}}^{\mathsf{\scriptscriptstyle T}}{\mathrm{A}}+\lambda\Phi$ and
${\mathrm{b}}_{j}{\mathrm{b}}_{j}^{\mathsf{\scriptscriptstyle
T}}+{\mathrm{c}}_{j}{\mathrm{c}}_{j}^{\mathsf{\scriptscriptstyle T}}$, for
$j\in{\mathcal{I}}^{{\scalebox{0.63}{$(\mathscr{P})$}}}$, can be calculated
only once in the optimization procedure. Note that since (L-)BFGS only employs
the gradients of $f_{n}$, it is preferred when
$n_{{\scalebox{0.5}{$\\!{\mathscr{P}}$}}}$ is large.

###  A-I Hyperparameter Tuning

In order to utilize Algorithm 1, we need a suitable stable kernel
$\mathbbm{k}$. The choice of kernel type is based on the shape and regularity
of the impulse response to be estimated. Once the type of kernel is decided,
we need to determine the regularization weight $\lambda$ and the
hyperparameters $\theta_{\mathbbm{k}}$ specifying kernel $\mathbbm{k}$.
Therefore, it is required to estimate the vector of hyperparameters
$\theta:=[\lambda,\theta_{\mathbbm{k}}]$ in a given set of admissible
hyperparameters $\Theta\subseteq{\mathbb{R}}^{n_{\theta}}$. To this end, we
employ a cross-validation approach with a Bayesian optimization heuristic
[59]. More precisely, we divide the index set of data into two disjoint sets,
${\mathcal{I}}_{\text{T}}$ and ${\mathcal{I}}_{\text{V}}$, which will be used
for training and validation, respectively. We define the model evaluation
metric $v:\Theta\to{\mathbb{R}}$ as the calculated prediction error on the
validation data, i.e., $v$ is defined as

$v(\theta)=\frac{1}{|{\mathcal{I}}_{\text{V}}|}\sum_{i\in{\mathcal{I}}_{\text{V}}}\big{(}y_{t_{i}}-{\mathrm{L}}^{\\!{\mathrm{u}}}_{t_{i}}({\mathrm{g}})\big{)}^{2},$
(109)

where ${\mathrm{g}}$ denotes the impulse response identified by the proposed
method using training data and hyperparameters $\theta$. Following this, the
vector of hyperparameters is estimated as

$\hat{\theta}:={\mathrm{argmin}}_{\theta\in\Theta}\ v(\theta).$ (110)

Since the model evaluation metric $v$ is a function of hyperparameters
$\theta$ in a black-box oracle form, we need to employ a Bayesian optimization
heuristic, such as GP-LCB, to solve optimization problem (110). This can be
done using Matlab’s bayesopt function [59].

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