Improving equi-integrability for a family $\mathcal F$ in $L^1(\Omega)$

Let $\mathcal F$ be a weakly compact subset of $L^1(\Omega)$. Dunford–Pettis theorem says that $\mathcal F$ is uniformly integrable. Also, by de la Vallée-Poussin theorem we can find an increasing convex function $\Phi:[0,\infty)\to[0,\infty)$ that is super linear, i.e. $$\frac {\Phi(t)} t\to \infty \quad\text{ as t\to\infty }$$ such that $\{\Phi(u):u\in\mathcal F \}$ is bounded in $L^1(\Omega)$.

Is it possible to assert further that we can find such $\Phi$ so that the family $\{\Phi(u):u\in\mathcal F \}$ is also equi-integrable?

Ideally I would also want $\Phi$ to be $C^1$ but I guess we can do that by smoothing out $\Phi$, if we manage to find one.

• Does "equi-integrable" mean the same thing as "uniformly integrable"? – Nate Eldredge Aug 8 at 13:19
• @NateEldredge Essentially yes, though I am using it to mean something slightly weaker. I merely want that $\sup_{\mathcal F} \int_S |\Phi(u(x))| dx<\varepsilon$ provided that $|S|<\delta$ for sufficiently small $\delta$. My definition of uniform integrability would need $\{\Phi(u):u\in\mathcal F \}$ to be bounded in $L^1$ as well. – BigbearZzz Aug 8 at 13:24
• Got it. I've seen that property called "uniformly absolutely continuous". – Nate Eldredge Aug 8 at 14:02
• @NateEldredge I've also seen some author use uniformly absolutely continuous. The funny part is that I ended up trying to prove the statement using uniform integrability anyway. – BigbearZzz Aug 8 at 15:45

Suppose $\Phi$ satisfies the assumptions of de la Vallée-Poussin theorem, I will prove that we can find $\Psi:[0,\infty)\to[0,\infty)$ (satisfying the same assumptions) such that $\{\Psi(u):u\in\mathcal F \}$ is uniformly integrable since this would imply equi-integrability as well.

I claim that, from a given $\Phi$, we can construct a strictly increasing $\Psi:[0,\infty)\to[0,\infty)$ which is convex and superlinear satisfying an extra assumption which is $$\frac {\Phi(t)}{\Psi(t)}\to \infty \quad\text{as}\quad t\to\infty.$$

Assuming that the claim is true, for any $\varepsilon>0$ we can find $T$ such that $\Psi(t)\le \varepsilon \Phi(t)$ whenever $t>T$. Take $M:=\Psi(T)$ and note that since $\Psi$ is strictly increasing, $\Psi(t)>M \iff t>T$. Then \begin{align} \sup_{u\in\mathcal F} \int_{\{ \Psi(|u|) >M \}} \Psi(|u(x)|)\,dx &= \sup_{u\in\mathcal F} \int_{\{ |u|>T \}} \Psi(|u(x)|)\,dx \\ &\le \varepsilon \sup_{u\in\mathcal F} \int_{\{ |u|>T \}} \Phi(|u(x)|)\,dx \\ &\le \varepsilon \sup_{u\in\mathcal F} \int_{\Omega} \Phi(|u(x)|)\,dx \\ \end{align} which proves the uniform integrability of $\{\Psi(u):u\in\mathcal F \}$.

Proving the claim is a bit longer, but essentially I constructed a piecewise-linear convex function $\Psi$ by interpolating some chosen points of $\log(\Phi)$ and some extra modification to make it convex.

If you see any error in my reasoning, please mention it so that I can attempt to fix it. Thank you very much.

WLOG, $$\mathcal{F}$$ is indeed a collection of nonnegative integrable functions.

As you pointed out, it suffices to show that

From a given function convex strictly increasing and superlinear $$\Phi$$, we can construct a strictly increasing $$\Psi:[0,\infty)\to[0,\infty)$$ which is convex and superlinear satisfying an extra assumption which is $$\frac {\Phi(t)}{\Psi(t)}\to \infty \quad\text{as}\quad t\to\infty.$$

Because $$\Phi$$ is a convex function on $$[0, \infty )$$, the limits $$\Theta(t) = \lim_{h \rightarrow 0^+} \dfrac{ \Phi(t+h)-\Phi(t)}{h}$$ exist for all $$t$$ and furthermore:

• $$\Theta$$ is increasing, hence measurable. And for all $$t$$,(by monotone convergence)
• $$\Phi(t)= \int_{0}^t \Theta(s)ds$$

Because $$\Phi$$ is superlinear and $$\Theta(s)$$ is increasing, $$\lim_{t \rightarrow \infty} \Theta(t) = +\infty$$

As $$\Theta$$ is positive ($$\Phi$$ is increasing), we can define the function $$\Psi$$ as follows: $$\Psi(t) = \int_{0}^t \sqrt{\Theta(s)}ds$$

$$\Psi$$ inherits naturally the required properties from $$\Phi$$ via the properties of $$\Theta$$ as mentioned above.