Here’s a proof that the Volterra operator is not normal which I thought was clever: Let be defined by

If then

so is a bounded operator. It follows that there exists a unique adjoint operator satisfying for all . Since has finite Lebesgue measure, , and thus

It follows that and hence by Fubini’s theorem,

from which we see that .

If with , then for any , Hölder’s inequality yields

so that is Hölder continuous. Moreover, the kernel of is and

so . It follows that is a Hilbert-Schmidt operator and therefore is compact. It follows then from the Fredholm alternative (see \texttt{http://bit.ly/1Uh8BgP}) that if , then is an eigenvalue of . That is, for all . If , then as we must have . By the fundamental theorem of calculus, , and so . Differentiating both sides of the eigenvalue equation yields , and hence for some . Since , we have , and therefore . It follows that the spectral radius of , , is zero.

If is a bounded operator on a Hilbert space and then

so

and hence . By symmetry, , whence . It follows that . Recall Gelfand’s formula for the spectral radius of a bounded operator :

If is a self-adjoint operator and then

which implies . By induction it follows that for all and hence

If is normal, then by induction , and as is self-adjoint, , from which we see that the spectral radius of a normal operator is equal to its operator norm.

Since the spectral radius of the Volterra operator is zero and its operator norm is positive, we conclude that is not a normal operator.

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