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?

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    $\begingroup$ @Surb: Together with $u_n \rightharpoonup u$ this implies $u_n \to u$. $\endgroup$ – gerw Jan 26 '17 at 18:09

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.


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.

  • $\begingroup$ Nice intuitive answer! This is awfully unrelated but I want to know how did you manage to format the text like that? I've never seen that blue-ish highlight before. $\endgroup$ – BigbearZzz Jan 27 '17 at 14:44
  • $\begingroup$ @BigbearZzz Thanks. I don't see the blue text, and didn't try to make anything blue. Maybe it is a bug with the website? You can make text colored within math mode with the \color{blue}{text} command. See here: meta.math.stackexchange.com/a/10116/3060 $\endgroup$ – Nick Alger Jan 27 '17 at 18:35
  • $\begingroup$ That's weird... Thank you anyway. $\endgroup$ – BigbearZzz Jan 27 '17 at 20:05
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    $\begingroup$ @BigbearZzz Oh I understand what you are talking about now. That is the "code" formatting. There is a button that does it in the post editor; the button looks like "{}". Also you can do it manually by putting 4 spaces in front of the text you want to format as code. $\endgroup$ – Nick Alger Jan 27 '17 at 21:35
  • $\begingroup$ I see. That's news to me! $\endgroup$ – BigbearZzz Jan 27 '17 at 22:07

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