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I need to find the spectrum of an operator $T: C([0,1]) \to C([0,1])$ defined by $(Tf)(t) = \int_0^t f(x) dx$. I know that the spectrum is the set of all values $\lambda$ such that $\lambda I - T$ is not invertible, but I'm not sure how to go about finding those values when integrals are involved. I've found several other posts about finding the spectrum of an integral operator here, but they all seem to involve functions of two variables, and that's throwing me off.

Any advice would be appreciated.

Stephen Montgomery-Smith
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Annapox
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  • Note that the Volterra operator is a Hilbert-Schmidt operator and therefore is compact. The Fredholm alternative yields that if $\lambda\in\sigma(T)$, $\lambda\ne 0$, then $\lambda$ is an eigenvalue of $T$. It can be shown that the only eigenvalue of $T$ is $0$, whence the spectrum of $T$ is ${0}$. – Math1000 Nov 23 '19 at 01:06
  • Neither Hilbert-Schmidt operators nor the Fredholm alternative has come up in class (this is the first I'm learning of them), so I'm not sure I'm allowed to use them. – Annapox Nov 23 '19 at 01:11

3 Answers3

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The resolvent $(T-\lambda I)^{-1}$ is obtained by solving for $g$ in the following: $$ g = (T-\lambda I)^{-1}f \\ (T-\lambda I)g= f \\ \int_0^x g(t)dt-\lambda g(x)=f(x) \\ \int_0^xg(t)dt-\lambda\frac{d}{dx}\int_0^x g(t)dt=f(x) \\ \frac{d}{dx}\int_0^xg(t)dt-\frac{1}{\lambda}\int_0^x g(t)dt=-\frac{1}{\lambda}f(x) \\ \frac{d}{dx}\left(e^{-x/\lambda}\int_0^xg(t)dt\right)=-\frac{1}{\lambda}e^{-x/\lambda}f(x) \\ e^{-x/\lambda}\int_0^x g(t)dt = -\frac{1}{\lambda}\int_0^x e^{-u/\lambda} f(u)du \\ \int_0^xg(t)dt=-\frac{1}{\lambda}\int_0^x e^{(x-u)/\lambda}f(u)du \\ g(x)=-\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x e^{(x-u)/\lambda}f(u)du $$ So, $$ (T-\lambda I)^{-1}f = -\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x e^{(x-u)/\lambda}f(u)du $$ and the spectrum is $\sigma(T)=\{0\}$.

Disintegrating By Parts
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Another approach is to use the spectral radius formula $r(T) = \lim_{n\to\infty} \|T^n\|^{1/n}$. By induction, you show that $$ T^n f(t) = \int_0^t \frac{(t-x)^{n-1}}{(n-1)!} f(x) \, dx .$$ So $\|T^n\| = 1/n!$, and $\|T^n\|^{1/n} \to 0$.

Stephen Montgomery-Smith
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The spectrum is $\{0\}$. To prove let $c \neq 0$ and this consider the equation $Tf-\frac 1 c f=g$ where $g \in C[0,1]$. If $h=Tf$ this becomes $h-\frac 1c h'=g$ or $h'-ch=-cg$. Hence $(e^{-cx}h(x))'=-ce^{-cx} g(x)$. Integrating we get $h(x)=e^{cx} [C-\int_0^{x}e^{-ct} g(t) dt]$ for some constant $C$. Since $h(0)=0$ we get $C=0$. Next $f =h'=(-e^{cx} \int_0^{x}e^{-ct} g(t) dt)'$. Now it is quite easy to see that $\|f\| \leq M \|g\|$ for a constant $M$ independent of $g$. We have proved that for every $g$ there is a unique solution $f$ for $Tf-\frac 1 c f=g$ and that that map $g \to f$ is continuous. This means that $T-\frac 1 c I$ has a bounded inverse whenever $c \neq 0$.

Kavi Rama Murthy
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