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I am assigned the following problem (a piece of Exercise 6.5 in Brezis's book):

Let $(\lambda_n)$ be a sequence of positive numbers such that $\lim_{n\to\infty}\lambda_n=+\infty$. Let $V$ be the space of sequences $(u_n)_{n\geq 1}$ such that $$\sum_{n=1}^\infty\lambda_n|u_n|^2<\infty.$$ The space $V$ is equipped with the scalar product $$((u,v))=\sum_{n=1}^\infty\lambda_nu_nv_n.$$ Prove that $V$ is a Hilbert space.

Here is the hint that Brezis provides for this part of the problem:

Let $T:V\to\ell^2$ be the operator defined by $$Tu=(\sqrt{\lambda_1}u_1,\sqrt{\lambda_2}u_2,\ldots).$$ Clearly $\|Tu\|_{\ell^2}=\|u\|_V$ for all $u\in V$, and $T$ is surjective from $V$ onto $\ell^2$. Since $\ell^2$ is complete, it follows that $V$ is also complete.

It seems clear to me that Brezis intends the reader to use the open mapping theorem; the map is obviously linear, and the hint points out that the map is an isometry, which proves both that it is bounded as well as injective. As the hint says, it is also clear that the map is surjective. At this point, all we have is a continuous bijection from $V$ to $\ell^2$. But applying the open mapping theorem, we have that the inverse map is also continuous, so that $V$ is isomorphically isometric to $\ell^2$, and because $\ell^2$ is complete, $V$ is complete.

Here's my problem: the open mapping theorem requires both spaces to be Banach spaces. To use this argument to prove that $V$ is complete is circular reasoning.

Am I misunderstanding the intended path of the argument? How would Brezis's hint be filled in to a complete argument correctly?

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I don't think he is using that theorem. See… – Poppy Jan 25 '14 at 23:52
Ah. We can just directly observe that $T^{-1}$ is continuous (evident in the same way that $T$ being continuous was), and skip the open mapping theorem. – learner Jan 25 '14 at 23:58
If you have isometry between two spaces and one of them is complete, then the other one is too. – Poppy Jan 25 '14 at 23:58
If I understand correctly, the key observation being that the inverse of an isometry is itself an isometry and therefore continuous, which is why we don't need the open mapping theorem. – learner Jan 26 '14 at 0:00
See this:… – Poppy Jan 26 '14 at 0:11

Indeed, in order to use the open mapping theorem we need complete spaces. Hence it's not the good way to show that $V$ is complete.

However, if $(v_n)_{n\geqslant 1}$ is a Cauchy sequence in $V$, then the sequence $(Tv_n)_{n\geqslant 1}$ is Cauchy in $\ell^2$ endowed with the usual norm, hence it converges to some $w$. Since $T$ is surjective, there exists some $v\in V$ such that $w=Tv$, hence $v_n\to v$ in $V$. The proof admits completeness of $\ell^2$, but we can prove completeness of $V$ in the same spirit of the proof of completeness of $\ell^2$.

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