My aim is to prove that in a Hilbert space, any sequence has a weakly convergent subsequence. To prove this, I'm trying to prove that:

In a Hilbert space, every bounded and closed set is weakly relatively compact

I have tried it via Banach-Alaoglu theorem but I find it difficult as this theorem applies to the dual space with the weak-* topology and not the weak one. Anyway I have the feeling that there is an easier way to prove it.

Any help would be highly appreciated. Thank you.

  • $\begingroup$ "any bounded sequence has a weakly convergent subsequence" $\endgroup$ – David Mitra Jan 19 '14 at 12:29
  • 1
    $\begingroup$ A (complex) Hilbert space is (anti) isomorphic to its dual. Doesn't that solve the problem? $\endgroup$ – Olivier Bégassat Jan 19 '14 at 12:29
  • $\begingroup$ @DavidMitra Thank you, typo corrected. $\endgroup$ – Simpy Jan 19 '14 at 12:32
  • $\begingroup$ @OlivierBégassat I'm trying to prove it with the theorems that I know. I don't know why a Hilbert space is isomorphic to its dual. Is there another way? $\endgroup$ – Simpy Jan 19 '14 at 12:34
  • $\begingroup$ Do you know Riesz representation for Hilbert spaces? $\endgroup$ – Olivier Bégassat Jan 19 '14 at 12:34

Banach-Alaoglu and Riesz theorem are an overkilling, even if I really like it as a proof. Moreover, your statement is true in general in reflexive spaces and it is usually known as the easy part of Eberlein-Smulian theorem. Recalling the following definitions,

Definition. Let $X$ be a Banach space. $X$ is reflexive iff the canonical injection $J: X \to X''$, $\langle J, x \rangle_{X'',X} = \langle \cdot, x \rangle_{X',X}$, is surjective.

A fundamental result is

Theorem 1 (Kakutani). Let $X$ be a Banach space. They are equivalent

  1. $X$ is reflexive;
  2. $B_X$ is weakyl compact.

From Kakutani's theorem follow

Theorem 2. Let $X$ be reflexive, $M$ a closed linear subspace of $X$. Then $M$ is reflexive.

Theorem 3. $X$ reflexive $\iff$ $X'$ reflexive.

Corollary 1. Let $X$ be reflexive and $M$ a closed linear subspace of $X$. Then $X/M$ (quotient space) is reflexive.

Corollary 2. $X$ reflexive and separable $\iff$ $X'$ reflexive and separable.

Now we can state

Theorem 4. Let $X$ be reflexive and $\{ x_n \} \subseteq X$ a bounded sequence. Then there exist $x \in X$ and $\{ x_{n_k}\}$ s.t. $x_{n_k} \rightharpoonup x$ as $k\to\infty$.

Proof. Set $M = \overline{\mbox{span}_{\mathbb K}}\{x_n\}$. By Theorem 2, $M$ is reflexive and obviously is separable. Hence, by Corollary 2, $M'$ is reflexive and separable. Then $J(B_M) = B_{M''}$ is compact and metrizable with respect to the $\sigma(M'', M')$ topology, so there exists $\{x_{n_k}\}$ subsequence and $F = J(x)$ s.t. $J(x_{n_k}) \rightharpoonup J(x)$ when $k\to \infty$ with respect to the $\sigma(M'', M')$, i.e., by definition, $\langle J(x_{n_k}), f \langle_{M'',M'} \to \langle J(x_n), f \rangle_{M'',M'}$. Since $M$ is closed, $x \in M$ and $\langle f, x_{n_k} \rangle_{M',M} \to \langle f, x_n \rangle_{M',M}$ when $k\to\infty$. Thanks to the Hahn-Banach theorem, there exists $L \in X'$ s.t. $f := L_{|M}$ and hence we have $\langle L, x_{n_k} \rangle_{X',X} \to \langle L, x_{n_k} \rangle_{X',X}.$ $\square$

Now consider a Hilbert space $H$. The scalar product on $H$ induces a norm which satisfies the parallelogramm identity (trivial):

$$\| x + y \|^2 + \| x - y \|^2 = 2( \|x\|^2 + \|y\|^2)$$

for all $x,y\in H$. Such a norm is uniformly convex and hence, by

Theorem 5 (Milman). Let $X$ be a Banach space. If $X$ is uniformly convex, then $X$ is reflexive.

we have $H$ is reflexive. So the previous Theorem 4 holds and your claim follows as a corollary of it.

I know your method was quicker, but I listed a few of results you would probably know, provided you can apply Banach-Alaoglou theorem. By the way, in the previous work there is no needing of knowledge of Hilbert spaces theory, except for the definition. In this sense, this is the easiest proof, even is longer.


  • Brezis, Functional analysis, Sobolev spaces and partial differential equations.
  • Yosida, Functional analysis.
  • $\begingroup$ Very Helpful! Thanks! $\endgroup$ – Xiao Jul 31 '14 at 15:25

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