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Let $H$ be an Hilbert space over $\mathbb{C}$

Let $V$ and $U$ be two closed subspaces of $H$

Let $\{v_m\}_{m \in \mathbb{N}} \in V$ be a basis for $V$

Let $\{u_m\}_{m \in \mathbb{N}} \in U$ be a basis for $U$

We know that $\{v_m\}_{m \in \mathbb{N}}$ is equivalent to $\{u_m\}_{m \in \mathbb{N}}$ that is: there exists a topological isomorphism $T : V \to U$ such that $\forall m \in \mathbb{N}: T(v_m) = u_m$

My question is if is it true that: $V+U$ is closed

Thanks

Matey Math
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2 Answers2

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I suppose you are talking about orthonormal bases. No infinite dimensional Hilbert space has a countable Hamel basis.

Any two separable infinite dimensional Hilbert spaces are isometrically isomorphic so your assumption of existence of a topological isomorphism is very weak. Any well known example of two infinite dimensional separable Hilbert subspaces whose sum is not closed would be a counterexample.

See Direct sum of two closed subspaces of Banach space is not closed

Kavi Rama Murthy
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  • True, but in this particular question we do not know in advance that the two bases ${v_m}$ and ${u_m}$ are orthogonal bases; therefore in order to complete you argument, you would have to show that any two bases for a separable Hilbert space are equivalent. Do you know this to be true? – uniquesolution Aug 26 '19 at 07:43
  • Yes, I know that every orthronormal ${u_m}$ can be isometrically transformed to another orthonormal basis ${v_m}$. But we are not given that the bases are orthonormal. What if the bases are not known to be orthonormal? How do you prove that there is an isomorphism taking one to the other? – uniquesolution Aug 26 '19 at 07:52
  • Regarding your edit right now: There is no reason to suppose that the bases are orthornormal! – uniquesolution Aug 26 '19 at 07:55
  • @uniquesolution His bases cannot be Hamel bases except when the subspaces are f.d.. If OP wants to talk about Scahuder bases he should mention that explicitly. I think it is reasonable to assume that his bases are orthonormal. – Kavi Rama Murthy Aug 26 '19 at 08:01
  • thansk @KaviRamaMurthy for your answer – Matey Math Aug 26 '19 at 10:25
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No. The sum need not be closed. Consider the example given here. We start with an orthonormal basis $\{e_n\}_{n=1}^{\infty}$ in $\ell_2$ (say) and define two subspaces as follows: $$M={\rm span}\{e_{2n}\}_{n=1}^{\infty},\quad N={\rm span}\{e_{2n}+\frac{e_{2n+1}}{n+1}\}_{n=1}^{\infty}$$ A general vector in $M$ has nonzero entries only at even indices, $(0,a_1,0,a_2,\dots)$, and a general vector in $N$ has the form $$(0,a_1,\frac{a_1}{2},a_2,\frac{a_2}{3},a_3,\frac{a_3}{3}\dots)$$ That $M+N$ is not closed was proved in the first accepted answer in the link above. So it remains to show that there exists a topological isomorphism from $M$ to $N$ carrying the basis of $M$ to the basis of $N$. Define $T:M\to N$ by the obvious choice: $$T(e_{2n})=e_{2n}+\frac{e_{2n+1}}{n+1}$$ Then a general vector $(0,a_1,0,a_2,\dots)$ in $M$ is transformed by $T$ into the general vector $x=(0,a_1,\frac{a_1}{2},a_2,\frac{a_2}{3},a_3,\frac{a_3}{3}\dots)$ in $N$. The norm of the first of $x$ is clearly $\|x\|=(\sum_{n=1}^{\infty}a_n^2)^{1/2}$, whereas the norm of the $Tx$ satisfies $$\|x\|\leq \|Tx\|=\left(\sum_{n=1}^{\infty}a_n^2+\sum_{n=1}^{\infty}\frac{a_n^2}{(n+1)^2}\right)^{1/2}\leq 2\|x\|$$ Since $\|x\|\leq\|Tx\|\leq 2\|x\|$, we see that $T$ is a topological isomorphism.

uniquesolution
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