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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.

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  • $\begingroup$ 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. $\endgroup$ – Robert Israel Dec 18 '14 at 18:22
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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.

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  • $\begingroup$ Thank you for your answer. Do you happen to have an idea how to adjust your example to tackle the edit? $\endgroup$ – Matthias Ludewig Dec 18 '14 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 Dec 18 '14 at 18:07
  • $\begingroup$ Sorry, I had a typo. But the other answer seems to solve the question! $\endgroup$ – Matthias Ludewig Dec 18 '14 at 20:00
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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.

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    $\begingroup$ 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. $\endgroup$ – Bill Johnson Dec 18 '14 at 19:43

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