A characterization for the derivative of a function being in $L^2([0,1]).$

Question

I'm preparing my PHD qualifying exam in functional analysis and measure theory. I looked at a problem from the previous exam and I got stuck trying to solve it

Let $F:[0,1]\rightarrow \mathbb{R}$ be a measurable function. Show that the following statements are equivalent:

1. There exists $f\in L^2([0,1])$ such that $$F(x)=\int_a^x f(t) dt.$$

2. There exists $M>0$ such that $$\sum_{k=1}^n \frac{|F(x_k)-F(x_{k-1})|^2}{x_k-x_{k-1}}\le M,$$

for each $n$ and every choice of points $x_0<x_1<\cdots x_n$.

So far, I've proven 1 implies 2. And for the converse I managed to prove that $F$ is absolutely continuous by taking $\varepsilon>0$, a finite set of intervals $(a_i,b_i)$ with lengths such that $\sum(b_i-a_i)<\delta$, with $\delta< \frac{\varepsilon^2}{M}$ and $i=1,\cdots,n$ and then noticing \begin{align*} \sum_{i=1}^n |F(b_i)-F(a_i)|&\le \sum_{i=1}^n \sqrt{M} \sqrt{b_i-a_{i}} \\ &\le \sqrt{M}\sqrt{\sum_{i=1}^n(b_i-a_i)} \\ &<\sqrt{M}\sqrt{\delta}\\ &<\sqrt{M}\sqrt{\frac{\varepsilon^2}{M}}\\ &=\varepsilon \end{align*}

I have that $F$ is absolutely continuous so I have the existence of a function $f\in L^1([0,1])$ that satisfies the equality in 1. But I can't imply that $f\in L^2([0,1])$.

In one of my attempts I used Lusin's Theorem and Tietze extension theorem to prove that there is a sequence $g_n$ of continuous functions with $g_n$ converging pointwise a.e. to $f$ with $f=g_n$ outside compact sets $K_n$ with $m(K_n)<\frac{1}{n}$ (or any positive bound that vanishes as $n\to\infty$). But when I try to find bounds for $\int_{K_n^C} f^2$ in I don't know what to do.

I also suspect there is a better approach to this problem. Something I must be missing out that would give us the condition $f\in L^2$ without the need to prove $f\in L^1$.

Thanks in advance for your help.

Answer 1

Suggestion:

This is for (2) implies (1). Once the absolute continuity of $F$ has been properly established and setting $f=F'$ (which exists a.s.) define

$$\begin{align} g_n(x)=\sum^{2^n-1}_{k=0}\Big(\frac{1}{2^n}\int^{x_{n,k+1}}_{x_{n,k}}\overline{f}\Big)\mathbb{1}_{(x_{n,k},x_{n,k+1}]}(x)f(x)\end{align}$$ where $\overline{f}$ is the complex conjugate of $f$, and $x_{n,k}=\frac{k}{2^n}$ for $k=0,\ldots, 2^n$. This corresponds sums of the type described in (2) fo the dyadic partition of the unit interval $[0,1]$.

For any Lebesgue point $x$ of $f$, one has that $$ g_n(x)\xrightarrow{n\rightarrow\infty}|f(x)|^2 $$

Integrability of $|f|^2$ follows from the uniform boundedness of $\{g_n\}$ and Fatou's lemma.