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Let $A: D(A) \subset X \rightarrow X$ be a generator of a $C_0-$semigroup and $Z$ be a bounded operator on $X$, then the evolution equation for $u \in C([0,T], \mathbb{R})$ $$\varphi'(t) = A \varphi(t) + Z u(t) \varphi(t)$$ with $\varphi(0)=\varphi_0 \in D(A)$ has a unique solution.

I would like to know if the following is true:

Let $\varphi_0 \in V \cap D(A)$ where $V$ is a closed subspace of $X$. If we have that $\langle \varphi'(t) , \psi \rangle =0$ for all $\psi \in V^{\perp}$ does this imply that $\varphi([0,T]) \subset V$?

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  • $\begingroup$ If $\varphi'(t)$ is orthogonal to $V$, why shoud $\varphi(t)$ stay in V? $\endgroup$
    – Jochen Wengenroth
    Commented Mar 16, 2017 at 13:47
  • $\begingroup$ @JochenWengenroth...you are right, a $\perp$ is missing. Otherwise, this even fails in $\mathbb{R}^n$ $\endgroup$
    – gipom
    Commented Mar 16, 2017 at 16:23
  • $\begingroup$ Is $N$ equal to $Z$ here? $\endgroup$
    – Phil Isett
    Commented Mar 16, 2017 at 16:49

1 Answer 1

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If $\varphi'(t)$ is orthogonal to $V^\perp$ then it belongs to $V^{\perp\perp}=V$, hence $$\varphi(s)=\varphi(0)+\int_0^s\varphi'(t)dt \in V.$$

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  • $\begingroup$ Good and concise answer that shows that no information about $A$ and $Z$ is needed, once well-posedness is established. Let me add that Jochen's answer uses the jargon of Hilbert space theory, but in fact a similar assertion also holds for general subsets $V$ of Banach spaces $X$: by the bipolar theorem $\phi(t)$ lies in the closed absolutely convex hull of $V$, which certainly agrees with (the closure of) $V$ if in particular $V$ is a (closed) subspace. $\endgroup$
    – Delio Mugnolo
    Commented Mar 28, 2017 at 6:22

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