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?
Solution 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.