Squeezing more convergence from the convergence in all $L^p$ spaces

Question

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?

Answer 1

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 $n\to\infty$.

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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.

Answer 2

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.

Answer 3

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 Meyer-Nieberg (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.