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?

  • $\begingroup$ The limits do not commute! $\endgroup$
    – Ian
    Feb 18, 2019 at 3:42
  • $\begingroup$ @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. $\endgroup$ Feb 18, 2019 at 3:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – Ian
    Feb 18, 2019 at 14:27
  • $\begingroup$ 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 $[0,1]$. $\endgroup$
    – MaoWao
    Feb 18, 2019 at 16:00
  • $\begingroup$ @MaoWao: yes; discovering that fact is why I asked this question. $\endgroup$ Feb 18, 2019 at 21:09

1 Answer 1


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 $$\epsilon>|g(0)-\langle g,\psi_{\delta}\rangle|$$

  • $\begingroup$ I see; there is a hidden uniformity assumption. Thank you! $\endgroup$ Feb 18, 2019 at 21:08

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