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let‎ ‎$ f \in L‎^{2} [‎ 0 ,‎ 1‎ ‎]‎ $‎.

‎define:‎ $ V( f ( t ) ) = ‎\int‎_{0}‎^{t} f‎ (‎ ‎s) ‎ds‎ $.(Volterra operator)

‎we ‎know ‎that‎ ‎$ ‎ V( f ( t ) ) ‎\in C‎ [‎ 0‎ ,‎ 1‎ ‎] ‎‎$ ‎for ‎all ‎‎$ f \in L‎^{2} [‎ 0‎ ,‎ 1‎ ‎]‎‎‎‎ $ ‎

can ‎we ‎find‎ ‎‎$ ‎\sigma (‎ V‎ )‎ =‎ ‎\{ 0‎ ‎\}‎ $ ‎and ‎$ r ( V ) = 0 $‎‎‎?‎‎

‎‎ ‎ we ‎have‎ ‎$ V‎^{*} :‎ ‎L‎^{2} [‎ 0‎ ,‎ 1‎ ]‎ ‎‎‎\longrightarrow L‎‎^{2} [‎ 0‎ ,‎ 1‎ ‎]‎ $,

‎so‎ ‎$ < f , V f > = < V‎^{*} f‎ ,‎ f‎ ‎>‎ $‎ ‎

and‎ ‎$ V‎^{*} (‎ f‎ (‎ t‎ ‎)) =‎ ‎‎\int‎_{‎t‎}‎^{‎1‎} f‎ (‎ ‎s) ‎ds‎ $‎‎

is ‎it ‎right ‎to ‎say‎ ‎$ V‎^{*} V‎ $‎ ‎is ‎compact ‎and ‎normal ‎?‎ ‎

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    Could you remind what you mean by normal? I forgot the definition – 2017-01-20
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    The compactness is ok since both operator are compact – 2017-01-20
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    I believe [this](http://math.stackexchange.com/questions/808176/how-to-show-that-the-volterra-operator-is-not-normal) shows that it is not normal. – 2017-01-20

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You can construct the resolvent $(V-\lambda I)^{-1}$ by solving for $g$, given $f$. This is a first order ODE that can be solved with an integrating factor. Assuming $\lambda \ne 0$, $$ Vg -\lambda g = f \\ Vg -\lambda \frac{d}{dx}Vg = f $$ The solution $g=R(\lambda)f$ is defined for all $\lambda\ne 0$ by $$ R(\lambda)f = -\frac{1}{\lambda}f-\frac{1}{\lambda^2}\int_{0}^{x}e^{(x-t)/\lambda}f(t)dt. $$ This operator is easily shown to be bounded on $L^2$. So $\sigma(V)=\{0\}$ because the spectrum is contained in $\{0\}$ and cannot be non-empty because $V$ is a bounded operator. $V$ cannot be normal because the norm and spectral radius are the same for a normal operator, and the spectral radius of $V$ is $0$ because $\sigma(V)=\{0\}$.

$V$ is compact. So $V^*V$ is compact. And $V^*V$ is selfadjoint and, hence, normal.