Let $X$ be a space endowed with a finite measure $m$. Let $f_n : \to \mathbb C$ be measurable functions such that $f_n \le 1$ for all $n$ and $f_n \to 0$ in every space $L^p (X)$ with $p \in [1, \infty)$. Is there even more hidden information about convergence that could be squeezed from this context? Can one get some even stronger convergence than the one already given?

1$\begingroup$ There are sequences of characteristic functions that converge in $L^p$ for $p\geq 1$ but do not converge in $L^{\infty}$. May I ask what kind of convergence are you specifically looking for? $\endgroup$– Onur OktayCommented May 9, 2022 at 13:16

$\begingroup$ I am not looking for anything specifically. I genuinely only wonder whether the convergence that I described above implies something even stronger, or whether at least it can be put in a more conceptual form than the one in my question. To say that something converges in all $L^p$ (with $p \ne \infty$) looks a bit like an unfinished hack. $\endgroup$– Alex M.Commented May 9, 2022 at 13:24

1$\begingroup$ It is hard to say something varnishingly useful in this generality. Undoubtedly these sort of sequences live in the metrizable locally convex space $\displaystyle \bigcap_{p\geq 1} L^p$, which contain Banach spaces like $L^{\infty}$, $L_{exp}$, and several similar Orlicz spaces. You may find the book by Bennett & Sharpley of interest, so here's my two cents: books.google.com/books?id=HpqF9zjZWMMC&pg=PA243 $\endgroup$– Onur OktayCommented May 9, 2022 at 14:12

$\begingroup$ @OnurOktay: my personal two cents guess would be that, with the OP's $L^\infty$ bound, you can actually get convergence in ANY Orlicz space $L_\Phi$. Sounds plausible? $\endgroup$– leo monsaingeonCommented May 9, 2022 at 19:34

$\begingroup$ @OnurOktay: wait... isn't there a Lebesgue dominated convergence theorem in Orlicz spaces? If so clearly convergence in $L^p$ gives pointwise a.e. convergence, and then via the uniform $L^\infty$ boud we should get $L^\Phi$ convergence? $\endgroup$– leo monsaingeonCommented May 9, 2022 at 20:54
3 Answers
Claim: for any (continuous) Young function $\Phi$ (with arbitrarily fast growth at infinity) we have strong convergence $f_n\to 0$ in the Orlicz space $L^\Phi$. More precisely, I claim that the Luxemburg norm $$ \f_n\_{L^\Phi}=\sup\limits_{k>0}\int_X \Phi(f_n(x)/k)\,dm(x) \to 0 $$ as $n\to\infty$.
Proof: it is known (see e.g. these notes by Christian Léonard, lemma 1.16) that convergence in the Luxemburg norm is equivalent to \begin{equation} \int_X \Phi(f_n(x)/k)\,dm(x) \to 0 \tag{*} \label{eq:star} \end{equation} for any fixed $k>0$. But since $f_n\to f$ in $L^p$ we have at lease pointwise convergence $f_n(x)\to 0$ for $m$a.e. $x$. Now because $\Phi$ is a nice Young function we have in particular $\Phi(f_n(x)/k)\to\Phi(0)=0$ a.e. Since we have in addition the uniform bound $\f_n\_{L^\infty}\leq M<+\infty$ we have as a consequence $\Phi(f_n(x)/k)\leq \Phi(M/k)$, and the latter constant is in $L^1(X,m)$ because the measure $m$ is finite. By the classical Lebesgue dominated convergence we conclude that the convergence \eqref{eq:star} holds and the claim follows.

$\begingroup$ in some sense the content of this statement is that the gap from Orlicz to uniform is really abrupt. This, by the way, reminds me of this recent question of mine, very much in the same spirit I believe mathoverflow.net/questions/420247/… $\endgroup$ Commented May 9, 2022 at 21:44

1$\begingroup$ The proof needs a minor revision, for $L^p$ convergence ensures only a.e.convergence of a subsequence. For instance, let $I_{j,k} = [2^{j}k,2^{j}(k+1))$ for $j\geq 1$ and $k=0,1,\dots, 2^j1$, $g_{j,k}$ be the characteristic function of $I_{j,k}$, and $f_n = g_{j,k}$ for $n=2^j+k$. $f_n\to 0$ in $L^p$ for all $1\leq p<\infty$, $(f_n)$ has a subsequence $\to 0$ a.e., but $f_n\not\to 0$ a.e. Also, perhaps you need $\inf$ in place of $\sup$ in the definition of the norm in the first display. $\endgroup$ Commented May 10, 2022 at 10:42

$\begingroup$ The revision required by Onur Oktay can be done by the simple observation that $\f_\\to 0$ if and only if every subsequence has a further subsequence along which this convergence holds. $\endgroup$ Commented May 10, 2022 at 12:11

$\begingroup$ @leomonsaingeon: I understand that the convergence that I have implies convergence in the Orlicz space as you have shown, but are the two convergences equivalent? Can I deduce convergence in all $L^p$ spaces ($p \ne \infty$) from the one in the Orlicz space, or is the latter weaker in general? $\endgroup$– Alex M.Commented May 10, 2022 at 14:49

$\begingroup$ @JochenWengenroth Sir, I appreciate your humorous comment, alas I do not have the authority to require anything. Had the International Society for Humor Studies accepted my plea to serve as a referee for their journal degruyter.com/journal/key/humr/html , perhaps I would have had a few things to require ruthlessly, although there is an intimidating opposition worldwide facebook.com/groups/reviewer2 . I'm here to share a few that I have, and perhaps get to learn more from the stars & superstars of their fields who stop by. $\endgroup$ Commented May 10, 2022 at 15:37
This is a comment, not an answer but the system won't let me. It's not clear to me what you are looking for but the convergence you describe is precisely that for the Mackey topology, i.e., the finest l.c. topology compatible with the duality between $L^\infty$ and $L^1$. The latter is also the finest such topology which agrees with one or all of your $L^p$ topologies (even with convergence in measure) on the $L^\infty$ ball.
Since I still can't comment, let me add the fact that the $L^\infty$ ball is trivially uniformly integrable and there is a ton of literature on the coincidence of various kinds of convergence for sequences with this property.
I think it might be worthwhile to mention a few observations in addition to the other answers:
For a sequence $(f_n)$ that satisfies $f_n \le 1$ for all $n$, convergence in $L^p$ for all $p \in [1,\infty)$ does not give you more than convergence for only one $p \in [1,\infty)$, since convergence with respect to one $p$norm is equivalent to convergence with respect to all $p$norms.
This, as well as the situation in Orlicz spaces (see this answer by leo monsaingeon), are special cases of a general result about Banach lattices that goes apparently back to Amemiya and can, for instance, be found in Theorem 2.4.8 of the book Banach Lattices by MeyerNieberg (link to zbMATH):
Theorem. Let $E,F,G$ be Banach lattices, where $F$ and $G$ have order continuous norm (the latter condition is, for instance, satisfied for $L^p$spaces for $p \in [1,\infty)$). If $E$ embeds (via lattice homomorphisms) into both $F$ and $G$ and is an ideal in both spaces, then on every order interval in $E$ the norm topologies induced by $F$ and $G$ coincide.
(And the topologies on the order interval induced by the weak topologies in $F$ and $G$, respectively, also coincide.)
If we apply the theorem to $F = L^p(\Omega,\mu)$ and $G = L^q(\Omega,\mu)$ (for a measure space $(\Omega,\mu)$ and $p,q \in [1,\infty)$) and to the order interval $[1,1]$ in $E = L^\infty(\Omega,\mu)$, we see that convergence of a sequence $(f_n)$ in $[1,1]$ with respect to the $p$norm is equivalent to convergence with respect to the $q$norm.
Of course, the same thing can also be deduced from the dominated convergence theorem and its partial converse, but I think it's nice to see that there is a general and abstract principle behind this.