# Why does weak-$L^2$ convergence not imply pointwise convergence for continuous functions?

This question shows that $$L^2$$ convergence does not show pointwise convergence, even when the functions involved are continuous. This strongly contradicts my intuition, because I thought that weak-$$L^2$$ convergence sufficed for the same.

What is wrong with the following proof?

Let $$\{\psi_\delta\}_{\delta}$$ be approximate identities and continuous $$\{f_n\}\rightharpoonup f_{\infty}$$ in $$L^2$$. We claim $$f_{\infty}(0)=\lim_{n\to\infty}{f_n(0)}$$.

To see this, fix $$\epsilon$$. There exists $$\alpha$$ such that for any $$\delta<\alpha$$ and continuous $$g$$, $$\epsilon>|g(0)-\langle g,\psi_{\delta}\rangle|$$ For such $$\delta$$, then, $$\epsilon+\langle g,\psi_{\delta}\rangle>g(0)$$ Now let $$g=|f_{\infty}-f_n|$$; we have $$\epsilon+\langle|f_{\infty}-f_n|,\psi_{\delta}\rangle>|f_{\infty}(0)-f_n(0)|$$ As $$\{f_n\}\rightharpoonup f_{\infty}$$, we have $$\lim_{n\to\infty}{\langle|f_{\infty}-f_n|,\psi_{\delta}\rangle}=0$$ for any $$\delta$$. So take $$n\to\infty$$; we obtain $$\epsilon>\limsup_{n\to\infty}{|f_{\infty}(0)-f_n(0)|}$$ Now take $$\epsilon\to0^+$$.

### Worked example

We can test this with the example from the cited question: in $$L^2([0,1])\cap C([0,1])$$, take $$f_n(x)=\ln{\!(n)}e^{-nx}$$ and $$\psi_{\delta}(x)=\frac{1}{\delta}\chi_{[0,1]}\left(\frac{x}{\delta}\right)$$. As $$n\to\infty$$, $$\|f_n\|_2=\ln{(n)}\sqrt{\frac{1-e^{-2n}}{2n}}\to0$$ so $$\{f_n\}_n\to0$$, and thus $$\{f_n\}_n\rightharpoonup0$$.

$$f_n(0)=\ln{(n)}$$, so we should expect $$\lim_{\delta\to0^+}{\int_0^1{f_n(x)\psi_{\delta}(x)\,dx}}=\ln{(n)}$$ On the other hand, weak convergence should give $$\lim_{n\to\infty}{\int_0^1{f_n(x)\psi_{\delta}(x)\,dx}}=\int_0^1{0\cdot\psi_{\delta}(x)\,dx}=0$$

Well, $$\int_0^1{f_n(x)\psi_{\delta}(x)\,dx}=\frac{\ln{(n)}(1-e^{-n\delta})}{n\delta}$$ Taking $$n\to\infty$$ does yield $$0$$, and $$\delta\to0^+$$ yields $$\ln{(n)}$$! What gives?

• The limits do not commute! – Ian Feb 18 '19 at 3:42
• @Ian: Yes. The idea here is that, by taking the $\delta$ limit after the $n$ limit, we "spread out" the values of $x$ that affect evaluating $f_n(x)$, so that the value $f_n(0)$ becomes "visible" to the $L^2$ norm. – Jacob Manaker Feb 18 '19 at 3:47
• Right, in the example they don't commute. At the level of this proof, that means that there are continuous functions with very bad modulus of continuity at $0$, which are arbitrarily hard to approximately evaluate at $0$ using the approximate identity $\psi_\delta$. This is why @gerw 's answer is correct. – Ian Feb 18 '19 at 14:27
• As a side remark: Even if all the functions are continuous, strong $L^2$ convergence does not implies pointwise convergence everywhere. Just think of $f_n(x)=x^n$ on $[0,1]$. – MaoWao Feb 18 '19 at 16:00
• @MaoWao: yes; discovering that fact is why I asked this question. – Jacob Manaker Feb 18 '19 at 21:09

## 1 Answer

The problem in your attempt is:

To see this, fix $$\epsilon$$. There exists $$\alpha$$ such that for any $$\delta<\alpha$$ and continuous $$g$$, $$\epsilon>|g(0)-\langle g,\psi_{\delta}\rangle|$$

This is not true. Instead, you have the following: For every $$g$$, there exists $$\alpha$$ such that for $$\delta < \alpha$$ you get $$\epsilon>|g(0)-\langle g,\psi_{\delta}\rangle|$$

• I see; there is a hidden uniformity assumption. Thank you! – Jacob Manaker Feb 18 '19 at 21:08