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.

  • $\begingroup$ Does "equi-integrable" mean the same thing as "uniformly integrable"? $\endgroup$ – Nate Eldredge Aug 8 '18 at 13:19
  • 1
    $\begingroup$ @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. $\endgroup$ – BigbearZzz Aug 8 '18 at 13:24
  • $\begingroup$ Got it. I've seen that property called "uniformly absolutely continuous". $\endgroup$ – Nate Eldredge Aug 8 '18 at 14:02
  • $\begingroup$ @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. $\endgroup$ – BigbearZzz Aug 8 '18 at 15:45

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.

  • $\begingroup$ Sorry it's been so long since you've answered my question, I've just seen it a moment ago. After reading it I have one question regarding what you meant by " And for all t,(by monotone convergence)" $$ \Phi(t)= \int_{0}^t \Theta(s)ds. $$ May I ask how is this the result of monotone convergence? $\endgroup$ – BigbearZzz 2 days ago
  • $\begingroup$ Sure. Because $\Theta$ is convex, $\forall t : h \mapsto \frac{\Theta(t+h)-\Theta(t)}{h}$ increasing on $\mathbb{R}_+$. Then, we use the equality $\left( \frac{1}{h} \int_{t}^{t+h} \Theta(s)ds \right)- \left( \frac{1}{h} \int_{0}^h \Theta(s)ds \right) = \int_{0}^t \frac{ \Theta(s+h)-\Theta(s)}{h}.ds $. $\endgroup$ – Taro NGUYEN 2 days ago

Here is what I have thought so far about this problem.

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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.