Why weak convergence doesn't imply convergence?

Question

Let $(H,\langle,\rangle)$ be a Hilbert space. We have that $u_n$ converges to $u$ weakly if $$\lim_{n\to \infty}\langle u_n,v\rangle=\langle u,v\rangle$$ for all $v\in H$. But why doesn't it converge strongly? Indeed, if $v=u_m$, then, $$\lim_{n\to \infty }\langle u_n,u_m\rangle=\langle u,u_m\rangle$$ for all $m$, and thus $$\lim_{m\to \infty }\lim_{n\to \infty }\langle u_n,u_m\rangle=\langle u,u_m\rangle\lim_{m\to \infty }\langle u,u_m\rangle=\langle u,u\rangle$$ therefore, $$\lim_{n\to \infty }\langle u_n,u_n\rangle=\langle u,u\rangle\implies \lim_{n\to \infty }\|u_n\|=\|u\|$$ and thus it converges weakly. What's wrong here?

Answer 1

You have used $$\lim_{m \to \infty} \lim_{n \to \infty} a_{n,m} = a$$ implies $$\lim_{n \to \infty} a_{n,n} = a.$$ However, this is not true.

Consider, e.g., $$a_{n,m} = \begin{cases} \pi & m < n, \\ \mathrm{e} &n < m, \\ 42 & m = n.\end{cases}$$

Edit: It is also enlightening to consider the most famous weakly convergent sequence. Let $\{u_n\}_{n \in \mathbb N}$ be an orthonormal system. Then, your $a_{n,m}$ satisfies $$a_{n,m} = \begin{cases} 0 & m \ne n, \\ 1 &n = m.\end{cases}$$ Again, $$\lim_{m \to \infty} \lim_{n \to \infty} a_{n,m} = \lim_{n \to \infty} a_{n,n}$$ fails blatantly.

Answer 2

This can be understood geometrically.

To makes things easier, specify the limit point to be zero: $u = 0$. The general case $u \neq 0$ follows from affine translation.

The set $$\{ u : \langle u, v \rangle \le \epsilon \}$$ describes a semi-infinite slab contained between two closely spaced parallel hyperplanes with normal vector $v$. The convergence $$\langle u_n, v \rangle \rightarrow 0$$ occurs successfully if for every slab width, nomatter how thin, a tail of the sequence is contained within the slab. I.e., the sequence is asymptotically "controlled" in the $v$-direction. The sequence converges weakly if this holds for every possible slab direction. I.e.,

for each direction v:
    for each width ε:
        A tail of the sequence is contained in the slab with 
        direction v and width ε.

In contrast, the sequence converges strongly if every scaled copy of the unit ball, $$\{ u : ||u|| \le \epsilon \},$$ nomatter how small, contains a tail of the sequence:

for each width ε:
    A tail of the sequence is contained within the ball of radius ε.

Strong convergence controls the convergence in all directions simultaneously.

In infinite dimensions a problem occurs since one can create a well-spaced infinite sequence without going anywhere, by cycling through the infinity of different dimensions: \begin{align*} u_1 &= (1,0,0,\dots) \\ u_2 &= (0,1,0,\dots) \\ u_3 &= (0,0,1,\dots) \\ \vdots & \end{align*}

The sequence converges weakly since every direction is controlled eventually, but it does not converge strongly since you have to wait infinitely long for every direction to be controlled.