Let $1 \leq p < \infty$ be fixed and let $\Omega \subseteq \mathbb{R}^n$ be open. Let $(Q_n)_{n \in \mathbb{N}}$ be a uniformly bounded family of operators on $L^p(\Omega)$, i.e. there exists $C>0$ such that $\|Q_n\| \leq C$ for all $n \in \mathbb{N}$.

Now suppose that for all $u \in L^p(\Omega)$, we have $Q_n u \longrightarrow u$ pointwise almost everywhere. Does this imply that $Q_n \longrightarrow \mathrm{id}$ in the strong operator topology, i.e. $Q_n u \longrightarrow u$ in $L^p(\Omega)$ for each $u \in L^p(\Omega)$?

I am looking for a proof or a counterexample. Does the answer depend on the choice of $\Omega$?

Edit: What if we additionally have $\|Q_n\| \longrightarrow 1$, or even $\|Q_n\| \leq 1+ \frac{C}{n}$ for some $C>0$? I forgot to ask about this additional condition in my first post. I am aware that this probably still does not fix the situation, but I didn't manage to construct an example from the answer given that satisfies this additional requirement.

  • $\begingroup$ By Fatou's lemma you can't have $Q_n u \to u$ pointwise a.e. if $\|u\| > 0$ and $\|Q_n\| < 1$, so the condition $\|Q_n\| \le C/n$ isn't going to work. $\endgroup$ Commented Dec 18, 2014 at 18:22

2 Answers 2


Let us consider the case $p=1$. Let $u \in L^1(\mathbb{R})$ be a positive function.

Define $f_n := \chi_{[n,n+1]}$ and $Q_nu := u + f_n \star u$, where $\star$ denotes the convolution.

Notice that, for any such $u$, we have $\|Q_nu - u\|_{L^1(\mathbb{R}^n)} = \|u\|_{L^1} > 0$ and that the sequence $(Q_n)$ satisfies :

  • $ \|Q_nu\|_{L^1} \leq 2 \|u\|_{L^1} $

  • $ (Q_nu - u)$ is a sequence of continuous functions which vanishes pointwise when $n$ goes to $+ \infty$ (if you further assume that $u$ is compactly supported, then for any $x \in \mathbb{R}, (Q_nu - u)(x) \equiv 0$ for $n$ big enough)

For bounded $\Omega$, replacing $f_n$ by (something like) $g_n := n\chi_{[0,\frac{1}{n}]}$ probably works, provided that you mind about the boundary.

Anyhow, your claim is morally equivalent to a stronger version of the dominated convergence theorem (without domination), so it was bound to fail.

Hope this was clear enough, don't hesitate to ask for details.

  • $\begingroup$ Thank you for your answer. Do you happen to have an idea how to adjust your example to tackle the edit? $\endgroup$ Commented Dec 18, 2014 at 17:14
  • $\begingroup$ Your second assumption (namely, that $\|Q_n\| \leq \frac{C}{n}$) is incompatible with $Q_n \to id$. Regarding the case where $\|Q_n\| \to 1$, I will think about it, but will not be able to answer agan before a few days, so if anyone has an idea right now... $\endgroup$
    – Hachino
    Commented Dec 18, 2014 at 18:07
  • $\begingroup$ Sorry, I had a typo. But the other answer seems to solve the question! $\endgroup$ Commented Dec 18, 2014 at 20:00

Suppose $p > 1$ and $\|Q_n\| \to 1$. Using Fatou's lemma for a lower bound and the norm for an upper bound, $\lim_n \|Q_n u\|_p = \|u\|_p$, and similarly $\lim_n \|(u + Q_n u)/2\|_p = \|u\|_p$. By Clarkson's inequalities, $\lim_n \|Q_n u - u\|_p = 0$, i.e. $Q_n \to \text{id}$ in the strong operator topology.

This still leaves the case $p=1$ open.

  • 3
    $\begingroup$ If $f_n \to f$ a.e. and $\|f_n\|_p \to \|f\|_p$ then $\|f-f_n\|_p \to 0$ for any $p<\infty$. This is problem 6.10 in Folland's Real Analysis and is in many other RA books. Clarkson is not needed for the proof. $\endgroup$ Commented Dec 18, 2014 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.