Let $V$ be an infinite-dimensional separable Hilbert space and let $V_f$ be a subspace of $V$ that is finite dimensional. It follows that $V_f$ is closed.
Is it true that $L^2(0,T;V_f)$ is closed as a set in $L^2(0,T;V)$?
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Use the description of elements in terms of limits of simple functions. – Martin Jun 25 '13 at 10:25
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Do you mean to write the limit point (that I want to show is in $L^2(0,T;V_f)$ for closedness) in terms of limit of simple functions? – michael_faber Jun 25 '13 at 10:53
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Elements of $L^2(0,T;V_f)$ are by definition limits of sequences of simple functions with values in $V_f$. It doesn't matter whether you compute their norms as elements of $L^2(0,T;V_f)$ or $L^2(0,T;V)$, so the embedding is in fact isometric. – Martin Jun 25 '13 at 10:57
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I'm sorry I don't see why that implies closedness. Appreciate it if you please elaborate a bit. – michael_faber Jun 25 '13 at 14:17
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@Should've been $V_f$. – michael_faber Jun 27 '13 at 17:25
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Consider map $$ I:L_2((0,T),V_f)\to L_2((0,T),V),\varphi\mapsto \varphi|^{V} $$ which assotiates to $\varphi$ the same function but with values in larger space i.e. in $V$. Obviously $I$ is linear. It is also isometric. Indeed $$ \Vert I(\varphi)\Vert_{L_2((0,T),V)} =\left(\int_0^1\Vert \varphi\Vert_V^2 dt\right)^{1/2} =\left(\int_0^1\Vert \varphi\Vert_{V_f}^2 dt\right)^{1/2} =\Vert\varphi\Vert_{L_2((0,T),V_f)} $$ Since $I$ is isometric it is bounded below, hence its image is closed. So $L_2((0,T),V_f)\equiv\operatorname{Im}(I)$ is closed in $L_2((0,T),V)$
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If $V_f$ has a different norm to $V$, do you think it still holds? Assuming $V_f \subset V$ is continuous – michael_faber Jul 09 '13 at 18:02
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it will hold if $$\exists c>0\quad\forall v\in V_f\quad \Vert v\Vert_V\geq c\Vert v\Vert_{V_f}$$ – Norbert Jul 09 '13 at 18:15
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@Norbert So the subspace is also a banach space. Do you know if there is an easier way to see this other than "closed subspace of Banach space is Banach"? – aere Sep 30 '13 at 14:05
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