In last lines in the image from Lawrece. C. Evans, Partial Differentail Equations, he states that in Hilbert space every bounded sequence contains a weakly convergent subsequence.
What is wrong in my counter example ?
Let $\mathcal H$ be an infinite dimensional Hilbert space with orthonormal basis $\{u_k\}_{k=1}^\infty$. Take them as a sequence. Now this sequence is bounded. I don't understand what subsequence of it may converge weakly and to what it may converge.
Best Answer
The sequence you gave converges weakly to $0$. Let $f$ be any bounded linear functional on $\mathcal{H}$. By the Riesz representation theorem, $f =\langle x, {-} \rangle$ for some $x \in \mathcal{H}$. Since $\{u_k\}_{k =1}^\infty$ is an orthonormal basis for $\mathcal{H}$, we have $x = \sum_{k=1}^\infty \alpha_k u_k$ for some scalars $\{\alpha_k\}_{k =1}^\infty$. Now
$$\lim_{n \to \infty} f(u_n) = \lim_{n \to \infty} \left\langle \sum_{k=1}^\infty \alpha_k u_k, u_n \right\rangle = \lim_{n \to \infty} \alpha_n = 0.$$
Edit: I assumed the scalars are real here, but this argument also works over $\mathbb{C}$ (just conjugate where appropriate).