# Is the integral operator $I: L^1([0,1])\to L^1([0,1]), f\mapsto (x\mapsto \int_0^x f \,\mathrm d\lambda)$ compact?

Is the integral operator $$I: L^1([0,1])\to L^1([0,1]), f\mapsto (x\mapsto \int_0^x f \,\mathrm d\lambda)$$ compact?

For $$I: L^p([0,1]) \to C([0,1])$$ with $$p\in (1,\infty]$$ this can be shown quite easily using Hölder's inequality and the compactness criterion in the Arzelà-Ascoli-Theorem (see also the first example on Wikipedia).

I wonder whether the statement holds also for $$p=1$$. However, we cannot just apply Arzelà-Ascoli here, because a bounded sequence $$(f_n)_{n\in \mathbb N}$$ in $$(L^1([0,1]), \lVert\cdot\rVert_{L^1})$$ is not necessarily equicontinuous (as shown here). Moreover, $$I: L^1([0,1]) \to C([0,1])$$ is not compact as discussed in the comments.

Edit: As answered in the comments, $$I: L^1([0,1])\to C([0,1])$$ is not compact. Therefore, I removed this aspect of the question.

• You should clarify what you meant by elementary: explain what your background is for example. Dec 28, 2021 at 17:33
• I simply do not have a large foundation of functional analysis; I haven't attended any course nor have I read a complete functional analysis book. Dec 28, 2021 at 17:35
• How do you define a compact operator then? Dec 28, 2021 at 17:36
• A compact operator is a linear operator that maps bounded sets in one metric space to relatively compact sets in the other metric space. Dec 28, 2021 at 17:37
• Ok. Well, a quick search on Volterra operator gives me this: johnthickstun.com/docs/volterra.pdf Dec 28, 2021 at 18:01

Yes, the operator is compact from $$L^1$$ to $$L^1$$. This follows from a duality argument:
It can be easily checked that the dual operator $$I': L^\infty([0,1]) \to L^\infty([0,1])$$ is given by $$(I'g)(x) = \int_x^1 g \; d\lambda \quad \text{for } x \in [0,1]$$ for all $$g \in L^\infty([0,1])$$, so $$I': L^\infty \to L^\infty$$ is compact by the Arzelà-Ascoli theorem. But a bounded linear operator is compact if and only if its dual operator is compact, so $$I: L^1 \to L^1$$ is compact, too.