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On Banach Space $C[0,1]$, T is a bounded linear operator and is defined by $Tf(x)=\int_0^xf(y)dy$, then how can I determine the spectrum of T? I was hinted to first show $T^nf(X)=\frac1{(n-1)!} \int_0^x(x-y)^{n-1}f(y)dy$

Mike
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    after proving that $\rho(A)=0$ it is remains to recall the inclusion $\sigma(A)\subset{z\in\mathbb{C}:|z|\leq\rho(A)}$ – Norbert May 06 '13 at 19:59

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Using your hint, $$ \|T^nf\|=\sup_x \left|\frac1{(n-1)!} \int_0^x(x-y)^{n-1}f(y)dy\right|\leq\frac{2^n}{(n-1)!}\,\|f\| $$ so $\|T^n\|\leq 2^n/(n-1)!$. Then $$ \mbox{spr}(T)=\lim_n\|T^n\|^{1/n}\leq\lim_n\frac2{[(n-1)!]^{1/n}}=0, $$ using that $(n!)^{1/n}\to\infty$.

So the spectral radius is $0$, and thus $\sigma(T)=\{0\}$.

Martin Argerami
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