Antiderivative as an integral operator from $L^2(0,2\pi)$ to $L^2(0,2\pi)$

Question

I am starting to study Functional Analysis on Hilbert Spaces and I am studying the following operator:

$$T:L^2(0,2\pi) \rightarrow L^2(0,2\pi) $$ where $$Tf:(0,2\pi) \rightarrow \mathbb{R} \\ \hspace{1.5cm} x \longrightarrow Tf(x)=\int^x_{0}f(y) \ dy \hspace{0.2cm} \text{for each } x \in (0,2\pi)$$

I have proved:

1. It is linear

   Now I use that the domain and codomain are Hilbert Spaces and therefore, by proving $||Tf|| \leq k \cdot ||f|| $ and lineality of T it follows that

2. It is bounded

3. It is continuous

My problems start when I try to:

1. compute the kernel and the range of the operator.
2. I don't know if it takes bounded sets to relative compact sets (precompact sets) (a set is relative compact or precompact if its closure is compact). I think that this is the same as wondering if the operator is compact.

What I have thought, done or tried:

1. $$\ker(T)=\{f \in L^2(0,2\pi) : Tf=0_{L^2(0,2\pi)}\} \\ =\{f \in L^2(0,2\pi) : \int^x_{0}f(y) \ dy=0_{L^2(0,2\pi)} \forall x \in (0,2\pi) \}=0_{L^2(0,2\pi)}$$

I am not sure about the last equality. Could be for example something like the Dirichlet function but with $g(x)=1$ if $x$ is rational and $g(x)=-1$ if $x$ is irrational?

Thanks to dafinguzman for the comment (I include it here since maybe it could be useful for someone else):

Concerning your Dirichlet type function: it is almost everywhere identical to the constant −1 function, so their integrals would be the same and it wouldn't be in the kernel.

2. I think that the range must be continuous functions or at least integrals of Lebesgue square integrable functions but I have started studying Lebesgue integration a week ago so i have no clue about it. Maybe, could we use the Lebesgue Differentiation Theorem in order to claim that the range is the set of functions of $L^2(0,2\pi)$ which are differentiable a.e in $(0,2\pi)$ and such that their derivative is f?

3. In order to deal with the last question I know from basic Topology that in finite Hilbert Spaces closed and bounded subsets are compact, therefore the image of a bounded set by a bounded operator must be relative compact. However, in non finite dimensions things are not so easy.

I have looked for a few equivalences and I tried to show that the closure of the image of a bounded set (which is bounded since T is bounded) in $L^2(0,2\pi)$ is limit point compact (A space X is said to be limit point compact if every infinite subset of X has a limit point in X).

I haven't been able to prove that. Maybe I am working in the wrong direction and I should look for a counterexample? And what do you think about the kernel and range?

Answer 1

This operator is known as the Volterra operator.

As Martin R pointed out, its kernel is trivial. The range can't be described in simpler terms than "antiderivatives of $L^2$ functions", which is a tautology.

The operator is compact. One way to show it is to apply a general theorem saying that all Hilbert-Schmidt operators are compact (as is gone here). Another way is to note that $\|f\|_2\le 1$ implies, for $s<t$, $$ |Tf(s)-Tf(t)| \le \int_s^t |f(x)|\,dx \le \sqrt{t-s}\|f\|_{L^2}\le \sqrt{t-s} $$ by the Cauchy-Schwarz inequality. Hence, the image of the unit ball under $T$ is an equicontinuous, uniformly bounded family of functions. By the Arzelà-Ascoli theorem, every sequence in this family has a uniformly convergent subsequence, hence $L^2$ convergent.