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Let $\left\{S(t)\right\}_{0 \leqslant t \leqslant \theta}$ be a strongly continuous semigroup of linear continuous operators in Hilbert space $H$, $S(0) = I$. Let $x$ be some element of $H$. Then its evolution on interval $[0,\theta]$ is described by equation $x(t) = S(t)x$. Let $G$ be a linear operator that sends trajectory $$Tx=\left\{ x(t) = S(t)x \mid 0 \leqslant t \leqslant \theta\right\}$$ of point $x \in H$ to point $GTx \in H$. $G$ has a meaning of observation operator. Define a linear operator $K$ in Hilbert space $H$ by the rule $Kx = GTx$.

I have to work with operator $K$. I want $K \in \mathcal{L}(H)$ but I don't know how to obtain it in terms of operators $T$ and $G$. My question is if it is possible to consider a set $TH = \left\{ Tx \mid x \in H \right\}$ as a linear subspace of some Hilbert space $Y$ so that $T \in \mathcal{L}(H,Y)$. If it is possible then we can just require $G \in \mathcal{L}(Y,H)$ to obtain $K \in \mathcal{L}(H)$.


Let $H = l_{2}$, $S(t)\left\{ x_{k} \right\}_{k=1}^{\infty} = \left\{ e^{-k^2t} x_k \right\}_{k=1}^{\infty}$. Then a set of trajectories can be considered as a linear subspace of functions $$ TH = \left\{ x(t,y) = \sum\limits_{k=1}^{\infty} e^{-k^2 t} x_{k} \sqrt{\frac{2}{\pi}} \sin( ky ) \mid \left\{ x_k \right\}_{k=1}^{\infty} \in l_2 \right\} $$ in $Y = L_{2} \left( [0,\theta] \times [0,\pi] \right)$ and we will have $T \in \mathcal{L}(H,Y)$. So the question is if it is true for any strongly continuous semigroup.

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When you say "Let $G$ be a linear operator", what do you mean by that? You can't say that an operator is linear without knowing the space on which it acts. – user53153 Jan 12 '13 at 23:03
Also, I don't quite understand your example. Why is $y$ present there? Why is the space called $TH$, as if it's the image of $H$ under $T$? Should $Y=L_2([0,\theta\times [0,\pi]])$ be $Y=L_2([0,\theta\times [0,\pi]];l^2)$, since $x(t,y)$ is not a scalar function, but takes values in $\ell^2$? – user53153 Jan 12 '13 at 23:18
@PavelM 1. $G$ takes trajectories as input, a set of trajectories is linear since $S(t)$ are linear operators. 2. Here $TH$ is a set of functions $(t,y) \mapsto x(t,y)$, $t \in [0,\theta]$, $y \in [0,\pi]$. It is the image of $H$ under $T$ if we replace $H$ by $L_{2}[0,\pi]$ (they are both separable Hilbert spaces, so we can do it: $x=\left\{ x_k \right\}_{k=1}^{\infty} \leftrightarrow x(y)=\sum_{k=1}^{\infty} x_k \sqrt{\frac{2}{\pi}} \sin(k y)$) – Nimza Jan 13 '13 at 12:53

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