]> 5.2 Representation of the orthoprojector

### 5.2 Representation of the orthoprojector

Another way of determining the approximants follows from the observation that any element of the subspace ${M}_{n}$ can be regarded as an output ${Y}_{n}$ of the observed linear system
$\left\{\begin{array}{ccc}\hfill ẋ\left(t\right)& \hfill =\hfill & Ax\left(t\right)\hfill \\ \hfill x\left(0\right)& \hfill =\hfill & b\in {\mathbb{ℂ}}^{n}\hfill \\ \hfill {Y}_{n}\left(t\right)& \hfill =\hfill & {c}^{\ast }x\left(t\right)={〈x\left(t\right),c〉}_{{\mathbb{ℂ}}^{n}}\hfill \end{array}\right\},$

where

$A=diag\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{n}\right\},\phantom{\rule{1em}{0ex}}\sigma \left(A\right)={\left\{{\lambda }_{k}\right\}}_{k=1}^{n}\subset {\Pi }^{-},\phantom{\rule{1em}{0ex}}{c}^{T}=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill \dots \hfill & \hfill 1\hfill \end{array}\right].$

From the orthogonal projection theorem we get

 ${∥f-{P}_{n}f∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}={min}_{b\in {\mathbb{ℂ}}^{n}}{∥f-{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}$ (5.1)

Observe that

$\begin{array}{cc}\hfill {∥f-{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}=& {〈f-{Y}_{n},f-{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}=\hfill \\ \hfill =& {〈f,f〉}_{{\text{L}}^{2}\left(0,\infty \right)}-{〈{Y}_{n},f〉}_{{\text{L}}^{2}\left(0,\infty \right)}-{〈f,{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}+{〈{Y}_{n},{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}=\hfill \\ \hfill =& {∥f∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}-2Re{〈f,{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}+{∥{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}.\hfill \end{array}$

The term ${∥{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}$ can be calculated as follows

${∥{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}={\int }_{0}^{\infty }{\left|{c}^{\ast }{e}^{tA}b\right|}^{2}dt={b}^{\ast }\left({\int }_{0}^{\infty }{e}^{t{A}^{\ast }}c{c}^{\ast }{e}^{tA}dt\right)b={b}^{\ast }Hb,$

where

$H={\int }_{0}^{\infty }\left[\begin{array}{c}\hfill {e}^{\overline{{\lambda }_{1}}t}\hfill \\ \hfill {e}^{\overline{{\lambda }_{2}}t}\hfill \\ \hfill ⋮\hfill \\ \hfill {e}^{\overline{{\lambda }_{n}}t}\hfill \end{array}\right]\left[\begin{array}{cccc}\hfill {e}^{{\lambda }_{1}t}\hfill & \hfill {e}^{{\lambda }_{2}t}\hfill & \hfill \dots \hfill & \hfill {e}^{{\lambda }_{n}t}\hfill \end{array}\right]dt={\left[\frac{-1}{\overline{{\lambda }_{i}}+{\lambda }_{j}}\right]}_{i,j=1,2,\dots ,n}={H}^{\ast }>0$

since the pair $\left(A,{c}^{\ast }\right)$ is observable. Furthermore, $H$ is a unique solution of the Lyapunov matrix equation

 ${A}^{\ast }H+HA=-c{c}^{\ast }$ (5.2)

and $H$ is the Gram matrix of the system ${\left\{{e}^{\overline{{\lambda }_{k}}\left(\cdot \right)}\right\}}_{k=1}^{n}\subset {\text{L}}^{2}\left(0,\infty \right)$.

We now show a way to determine the scalar product ${〈f,{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}$.

$\begin{array}{cc}\hfill {〈f,{Y}_{n}〉}_{{\text{L}}^{2}\left(0,\infty \right)}=& {\int }_{0}^{\infty }f\left(t\right)\overline{{Y}_{n}\left(t\right)}dt={\int }_{0}^{\infty }f\left(t\right)\overline{{c}^{\ast }{e}^{tA}b}dt={\int }_{0}^{\infty }f\left(t\right){b}^{\ast }{e}^{t{A}^{\ast }}cdt=\hfill \\ \hfill =& {b}^{\ast }\left({\int }_{0}^{\infty }f\left(t\right){e}^{t{A}^{\ast }}dt\right)c={b}^{\ast }\left({\int }_{0}^{\infty }f\left(t\right)diag\left\{{e}^{\overline{{\lambda }_{1}}t},{e}^{\overline{{\lambda }_{2}}t},\dots ,{e}^{\overline{{\lambda }_{n}}t}\right\}dt\right)c=\hfill \\ \hfill =& {b}^{\ast }diag\left\{\stackrel{̂}{f}\left(-\overline{{\lambda }_{1}}\right),\stackrel{̂}{f}\left(-\overline{{\lambda }_{2}}\right),\dots ,\stackrel{̂}{f}\left(-\overline{{\lambda }_{n}}\right)\right\}c,\hfill \end{array}$

where $\stackrel{̂}{f}$ denotes the Laplace transform of $f$,

$\stackrel{̂}{f}\left(s\right)={\int }_{0}^{\infty }f\left(t\right){e}^{-st}dt.$

In virtue of the Paley–Wiener theory, $f\in {\text{L}}^{2}\left(0,\infty \right)$ iff $\stackrel{̂}{f}$ belongs to ${\text{H}}^{2}\left({\Pi }^{+}\right)$, the Hardy space of functions $\varphi$ analytic on the right complex half–plane ${\Pi }^{+}=\left\{s\in \mathbb{ℂ}:\phantom{\rule{0ex}{0ex}}Res>0\right\}$, such that

${sup}_{x>0}{\int }_{-\infty }^{\infty }{\left|\varphi \left(x+iy\right)\right|}^{2}dy<\infty .$

Since $\sigma \left(-{A}^{\ast }\right)$ is located in ${\Pi }^{+}$, the domain of analyticity of $\stackrel{̂}{f}$, we have (see [57, Theorem 5.3.2])

$\begin{array}{cc}\hfill {\int }_{0}^{\infty }f\left(t\right){e}^{t{A}^{\ast }}dt=& {\int }_{0}^{\infty }f\left(t\right)diag\left\{{e}^{\overline{{\lambda }_{1}}t},{e}^{\overline{{\lambda }_{2}}t},\dots ,{e}^{\overline{{\lambda }_{n}}t}\right\}dt=\hfill \\ \hfill =& diag\left\{\stackrel{̂}{f}\left(-\overline{{\lambda }_{1}}\right),\stackrel{̂}{f}\left(-\overline{{\lambda }_{2}}\right),\dots ,\stackrel{̂}{f}\left(-\overline{{\lambda }_{n}}\right)\right\}:=\stackrel{̂}{f}\left(-{A}^{\ast }\right).\hfill \end{array}$

Finally,

${∥f-{Y}_{n}∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}={∥f∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}-2Re\left[{b}^{\ast }\stackrel{̂}{f}\left(-{A}^{\ast }\right)c\right]+{b}^{\ast }Hb$

and the minimal value in (5.1) is achieved on $b\in {\mathbb{ℂ}}^{n}$ being a solution of the equation

 $Hb=\stackrel{̂}{f}\left(-{A}^{\ast }\right)c$ (5.3)

If ${b}_{0}$ is the solution of (5.3) then

 ${∥{P}_{n}f∥}_{{\text{L}}^{2}\left(0,\infty \right)}^{2}={b}_{0}^{\ast }H{b}_{0}={b}_{0}^{\ast }\stackrel{̂}{f}\left(-{A}^{\ast }\right)c$ (5.4)