Let $V$,$U$ and $W$ be closed and infinite dimensional subspaces of $\ell^2$ such that $V$ is isomorphic to $U$ and $W$
My question: Is $V$ isomorphic to $U + W$
Thanks.
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What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"? – GEdgar Jul 22 '18 at 19:02
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@GEdgar yes it is – Matey Math Jul 22 '18 at 19:03
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By isomorphic you mean isometrically isomorphic, right? – mechanodroid Jul 22 '18 at 19:33
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@mechanodroid yes it is – Matey Math Jul 22 '18 at 19:34
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No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)
However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $\ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V \cong \ell^2 \cong W + U$.
Nik Bren
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In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $\ell^2$. This is true for closed $\infty$-dimensional subspace of $\ell^2$. So the answer must be yes now. – Nik Bren Jul 22 '18 at 19:18
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thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered – Matey Math Jul 22 '18 at 19:20
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No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: https://math.stackexchange.com/questions/135471/the-direct-sum-of-two-closed-subspace-is-closed-hilbert-space/135547#135547 – Nik Bren Jul 22 '18 at 19:28
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The answer to the question as stated is no.
Let $V = \overline{\operatorname{span} \{e_{2n}\}_{n\in\mathbb{N}}}$ and $U = \overline{\operatorname{span} \left\{e_{2n} + \frac{e_{2n+1}}{n+1}\right\}_{n\in\mathbb{N}}}$.
Then $U, V$ are infinite-dimensional closed subspaces of $\ell^2$ and $V \cong V$, $V \cong U$. The latter are isometrically isomorphic via $$e_{2n} \mapsto \frac1{\sqrt{1+\frac1{(n+1)^2}}} \left(e_{2n} + \frac{e_{2n+1}}{n+1}\right)$$
However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.
mechanodroid
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