# An $L^1$ function but (really) no better?

Question: For a smooth, bounded domain $$\Omega\subset \mathbb R^d$$, does there exist a function $$u\in L^1(\Omega)$$ such that $$u\not\in L^\Phi(\Omega)$$ for any Orlicz space $$\Phi$$?

For the definition of Orlicz spaces see this wikipedia page. In a nutshell, a Young function is a convex function $$\Phi:\mathbb R^+\to \mathbb R^+$$ which is superlinear at infinity and sublinear at the origin, $$\frac{\Phi(u)}{u}\xrightarrow[u\to 0]{ }0 \quad\mbox{and}\quad \frac{\Phi(u)}{u}\xrightarrow[u\to +\infty]{ }+\infty$$ and (with a slight abuse) $$u\in L^\Phi \quad \mbox{iff}\quad \Phi(|u|)\in L^1.$$ The choice $$\Phi(u)=u^p$$ leads to the usual Lebesgue spaces $$L^p(\Omega)$$, so in some sense the Orlicz spaces allow to measure integrability on an arbitrary scale (instead of just pure powers). In particular, choosing $$\Phi$$ carefully allows somehow to measure fine regularity scales between $$L^1$$ and any other $$L^p$$ space. The archetypical and popular example is $$L^1\log L^1$$, corresponding to the choice $$\Phi(u)=u[\log u]^+$$

Just a few thoughts (before my question is abruptly downvoted and deemed inappropriate for MO): Of course for a given regularity scale, i-e for any fixed $$\Phi$$, it is easy to cook-up a function $$u=u_\Phi$$ that belongs to $$L^1$$ but not to $$L^\Phi$$ (just as one can easily tweak the negative exponents $$\alpha=\alpha(d,p)<0$$ so that $$u(x)=|x|^\alpha$$ is in $$L^1$$ but not in $$L^p$$ for fixed $$p>1$$ in dimension $$d$$). The question is therefore: can a (necessarily pathological) function $$u$$ be $$L^1$$ but really no better, i-e $$u$$ does not belong to any better $$L^\Phi$$ space? (I would perhaps call that such a function "essentially $$L^1$$".) Again, it is easy to construct $$u\in L^1$$ that does not belong to $$L^p$$ for any $$p>1$$, but typically this function might belong to $$L^1\log L^1$$. And if it doesn't then it might belong to $$L^1\log\log L^1$$, and so on... The game is: can one "exhaust all the regulatity scales" except for the minimal $$L^1$$ requirement? This seems highly unplausible, as the set of all possible regularity scales is highly uncountable, but who knows, some hidden monotonicity might save the day (going "down the scales"?)

Following the reverse line of thoughts, another related question is: if $$u\in L^1$$, can one always find some $$\Phi=\Phi_u$$ such that $$u\in L^\Phi$$ is in fact better than merely integrable? My intuition is that this should be blatantly false, but I realized that this is vaguely similar in spirit to Lusin's theorem, stating that essentially a measurable function is continuous. So perhaps this second counterintuitive statement of mine holds in some sense?

Some context: it turns out that for one of my research problems I am given a family $$(u_\xi)_{\xi\in G}\in \mathcal P(\Omega)$$ of Borel probability measures over $$\Omega$$. For $$\xi$$ in a smaller (but uncountable) set $$F\subset G$$ I know that $$u_\xi=u_\xi(x) \mathcal L(dx)$$ is in fact absolutely continuous w.r.t. the Lebesgue measure $$\mathcal L|_\Omega$$, and I would like to conclude (by a very intricate argument that I shall not discuss here) that this is the case also for $$\xi\in G$$. What I need for my argument to work is to find first some $$\Phi$$ such that $$u_\xi\in L^\Phi$$ somehow uniformly in $$\xi\in F$$, and then I would be able to control what happens for $$\xi \in G$$ using the same scale $$\Phi$$. Somehow, what I am trying to do is turning a mere pointwise absolute continuity into a globally quantified absolute continuity (the quantification being measured on a putative $$L^\Phi$$ scale). I am fully aware that at first glance this seems hopeless, but thinking again about Lusin's theorem gave me a vague hope. In any case, I would equally love to be proven wrong or to get some positive insight (a reference, better still).

There is a much more general result of Vallée-Poussin from which a negative answer to your question follows.

Let $$(X,\mu)$$ be a measure space. We say that a family of function $$\mathcal{F}\subset L^1(X)$$ is equi-integrable if for every $$\varepsilon>0$$ there is $$\delta>0$$ such that $$\sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta.$$

Theorem (de la Vallee Poussin). Let $$(X,\mu)$$ be a measure space with $$\mu(X)<\infty$$ and let $$\mathcal{F}\subset L^1(X)$$. Then $$\mathcal{F}$$ is equi-integrable if and only if there is a Young function $$\Phi$$, $$\lim_{t\to\infty}\Phi(t)/t=\infty$$ such that $$\sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1.$$

Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $$f\in L^1$$ there is $$\Phi$$ with $$\Phi(f)\in L^1$$.

Also we can adjust the function $$\Phi$$ around zero arbitrarily without changing integrability of the function $$\Phi(f)$$ so you can have the condition $$\Phi(t)/t\to 0$$ as $$t\to 0^+$$.

You can find the proof of the de la Vallee Poussin theorem in

C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.

M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.

• Oh yes of course, how stupid of me, how could I not see de la Vallée Poussin there? Thank you Piotr, much appreciated. Apr 12, 2022 at 20:49