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?

Question

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.

Answer 1

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.