1
$\begingroup$

Prove that if $x_n$ is a sequence in $H$ (Hilbert space) with $\sup_n||x_n||\le1$, then there is a subsequence $\{x_{n_j}\}$ and an element $x$ of $H$ with $||x||\le 1$ such that $x_{n_j}$ converges to $x$ weakly.

We did it for separable Hilbert spaces using orthonormal bases and Riesz Representation theorem. Does the argument simply extend to uncountable orthonormal basis in some easy way?

Also, how do we know the limit has norm 1? By continuity of the norm function?

$\endgroup$
1
$\begingroup$

Assume we know the result for separable Hilbert spaces. Then define $V$ as the closure of the vector space generated by the $x_j$, $j\geqslant 1$. We endow this space with the inner product of $H$: this is a separable Hilbert space. We extract a subsequence which converges weakly in $ V$. This means that for each $v\in V$, $$\tag{1}\lim_{k\to +\infty}\langle x_{n_k},v\rangle=\langle x,v\rangle.$$ We want to extend (1) to any $v\in H$. To this aim, since $V$ is closed, we may write $H=V\overset{\perp}\oplus V^{\perp}$. Then for $v\in H$, we may express it as $v'+v''$, where $v'\in V$ and $v''\in V^{\perp}$. Then $$\langle x_{n_k},v\rangle=\langle x_{n_k},v'\rangle +\langle x_{n_k},v''\rangle$$ and the second term is $0$ for each $k$ because $v''$ belongs to the orthogonal of $V$.

We can prove that in general, $$\lVert x\rVert\leqslant\liminf_{k\to\infty}\lVert x_{n_k}\rVert.$$ Indeed, since $\langle x,x\rangle=\lim_{k\to\infty}\langle x_{n_k},x\rangle$ we obtain by Cauchy-Schwarz inequality the wanted result. It's a kind of semi-continuity.

$\endgroup$
  • $\begingroup$ How to we know that the inner product converges for things in $V^{\perp}$ though? $\endgroup$ – Toeplitz Apr 28 '14 at 20:37
  • $\begingroup$ I've edited. Is it clearer? $\endgroup$ – Davide Giraudo Apr 28 '14 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.