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.
 A: 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.
A: 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.
