Is the antiderivative of an $L^2$ function always well-defined?

Question

Given $f$ in $L^2$ over some compact interval in the real line, I would like to define the antiderivative in the usual sense: $F(x)$ as the integral of $f$ from some constant up to $x$. I suppose this is not always possible; if not, what is the next best thing available? Can I always find $F(x)$ such that $F'= f$ in the weak/distribution sense? If so, what does $F$ look like?

Answer 1

In short, yes, we have an antiderivative.

Assuming you are familiar with absolute continuity of measures and some generalizations of the derivative, like the Radon-Nikodym Theorem, we proceed as follows. Let $\mathcal{B}$ be the standard borel-sigma algebra on $\mathbb{R}$ and $\mu$ denote Lebesgue measure on $\mathbb{R}$, and suppose that $f\in L(\mathbb{R},\mathcal{B},\mu)$ is a Lebesgue integrable function. We can define

\begin{equation} F: \mathcal{B}\rightarrow\mathbb{R}, \quad E\mapsto \int_E f d\mu \end{equation}

Per this result, the Lebesgue Integral can be used to define an absolutely continuous antiderivative, and by Radon-Nikodym, this function is differentiable $\mu$-a.e. Moreover, this derivative is unique up to equivalence a.e., so the derivative is $\mu$-a.e. equal to our original function $f$.

If you are only familiar with absolute continuity of functions and less general versions of Radon-Nikodym, then we can define another function

\begin{equation} G: \mathbb{R}\rightarrow\mathbb{R}, \quad E\mapsto \int_{[0,x]} f d\mu \end{equation}

and apply the weaker versions of the above theorems. Notice that $G$ is just $F$ composed with the map $\mathbb{R}\rightarrow\mathcal{B}, x\mapsto [0,x]$ so really these are saying the same thing, however I mention it since some courses only discuss absolute continuity of functions and not the more general notion.

Answer 2

I guess this answer has essentially the same content as the other one, but it is more succint.

If $f\in L^2([a, b])$ then the integral $$\tag{1} F(x):=\int_a^x f(y)\, dy $$ is well-defined for all $x\in[a, b]$. Indeed, by Cauchy--Schwarz, $$ \int_a^x \lvert f(y)\rvert\, dy\le \sqrt{\int_a^x\lvert f(y)\rvert^2\, dy}(x-a)^\frac12\le \lVert f\rVert_2(b-a)^\frac12,$$ and the right-hand side is finite. So the integral in (1) is absolutely convergent for all $x$.

Essentially the same proof shows that $F$ is $\frac12$-Hölder continuous. Moreover, $F$ is differentiable for almost all $x\in[a, b]$, which is not really important. Much more importantly, $F$ defines a distribution and its distributional derivative $F'$ equals $f$.

Sketch of proof of the latter statement: for all $\phi\in C^\infty_c((a, b))$, $$ -\int_a^b F(x)\phi'(x)\, dx = -\int_a^b \int_a^x f(y)\phi'(x)\, dydx = \int_a^b f(x)\phi(x)\, dx.$$