Uniform Integrability and relative compactness

Question

I am trying to proof relative compactness in L2(0,1) for a specific set of functions $(\phi_n)_{n \in \mathbb{N}}$ with following properties:

1. $\int_0^1 \phi(x) dx = 0 $
2. $||\phi_n^2||_{L1(0,1)} = 1 $
3. $(\phi_n)_{n \in \mathbb{N}}$ is uniformly integrable such that for each $\epsilon >0$ there exist a $\delta> 0$ fulfilling: $$\sup_{n \in \mathbb{N}} \int_0^1 \phi_n(x)^2*\mathbb{1}\left(|\phi_n(x)| > \delta\right) dx < \epsilon $$ Here, $\mathbb{1}$ is the indicator function resulting in 1 if the inner statement is true.
4. Each $\phi$ is non decreasing and has at most countable many discontinuity points. (Otherwise it is continuous)

I tried to use the theorem of Kolmogorow-Riesz, where I have to proof boundedness in the norm (which is already given above) and equicontinuity: $$ \lim_{h \rightarrow 0} \sup_{n \in \mathbb{N}} \int_0^1 \left(\phi_n(x+h) - \phi_n(x)\right)^2 dx$$

But I have difficulties to show the equicontinuity, since I do not find a possibility to use the given uniform integrability. I plotted many examples and I think that this property is necessary in this scenario. Without this property I think there could be a $\phi_n$ that has all its mass at the border of 0 and 1 so that the equicontinuity can not hold anymore. I Imagine that the uniform integrability should lead to distributing the mass uniformly so that the actual difference of $\phi_n(x+h)-\phi_n(x)$ is bounded in terms of h.

Does anyone have a source where something like this is done or is there a more elegant way than Kolmogorow-Riesz for showing the relative compactness?

Answer 1

Assume first that the $\phi_n$ are uniformly bounded in absolute value. Otherwise replace them by $K\mathbb{1}_{\phi_n\geq K}+\phi_n\mathbb{1}_{|\phi_n|<K}-K\mathbb{1}_{\phi_n<-K}$. We can extract a subsequence which converges for every rational number. This yields a non-decreasing function $f:\mathbb{Q}\cap [0,1]\to \mathbb{R}$. Extend it to $[0,1]$ by defining $f(x)=\lim_{t\to x^+, t\in \mathbb{Q}}f(t)$, where the limit exists by monotonicity. This is again non-decreasing, and so by a standard result it has only countable many discontinuity points. But note that for any continuity point, the left and right limits among the rationals agree. Hence, by monotoncity, it follows that at continuity points along the subsequence above, it holds $f(x)=\lim_{n \to \infty}\phi_n(x)$. Indeed, fix $s<x<t \in \mathbb{Q}$ close to $x$ so that $|f(x)-f(s)|<\epsilon$, and similarly for t. Then, for any n, $\phi_n(t)\geq \phi_n(x) \geq \phi_n(s)$, so both the limsup and liminf of $\phi_n(x)$ will be at most distance $\epsilon$ from $f(x)$, so that since $\epsilon$ was arbitrary, we indeed get the required convergence. Hence, we indeed get the convergence almost everywhere (since discontinuity points are countable) when the $\phi_n$ are uniformly bounded. In general, we can also vary the $K$ we used above to get the uniform boundedness, and extract via another diagonal argument another subsequence that converges almost everywhere on the set where the $\sup_n|\phi_n|<\infty$. From here, compactness follows from the fact that uniform integrability together with convergence almost everywhere implies convergence in norm.