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Consider $X=L^2[0,d]$ and define $A \colon X \to X$ a linear bounded operator by the integral function \begin{equation} {A}x=\int_0^\cdot x(\xi)d\xi \end{equation} This operator looks compact to me since its range $R(A) \subset W^{1,2}$ which is continuously embedded with compact embedding into $X$. Is this true? How can I formalize the details?

carlos85
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  • This is the Volterra operator. One way to see that $A$ is compact, is to realise that it is a Hilbert--Schmidt operator. A more direct way to prove compactness is to apply the Arzela--Ascoli theorem, see https://math.stackexchange.com/questions/151425/spectrum-of-indefinite-integral-operators. – Janik Mar 04 '22 at 15:55
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    Your idea is correct: you need to prove that $A$ is continuous from $L^2$ to $W^{1,2}$ to finish the proof. – daw Mar 04 '22 at 15:57

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