Point Spectrum, Spectrum for Operator

Question

I am having trouble finding the spectrum and point spectrum of the following operator.

$T:C[0,1] \to C[0,1]$

$Tf(x) = \int_0^xf(t)dt \quad x\in[0,1]$

Any help would be appreciated

Answer 1

Suppose we had a non-zero $\lambda$ in the point spectrum. What does it mean for something to be in the point spectrum? It means that $$(T-\lambda)f=0.$$ That is, $$\int\limits_0^x f(s)\, ds=\lambda f(x).$$ Note that, via the FTC, $T$ sends $f$ to a differentiable function. If we differentiate the above, we have the differential equation $f=\lambda f'.$ This has solutions of the form $f(x)=ce^{x/\lambda}$ (which needs that $\lambda\neq 0)$. If we evaluate this at $x=0,$ we get that $f(0)=c.$ However, we can see from our integral equation that at $x=0,$ $\lambda f(0)=0.$ So, if $\lambda\neq 0,$ then $c=0,$ implying $f\equiv 0.$ So, there is no non-zero point spectra. The $\lambda=0$ case is trivial. I'll leave it to you to discuss the general spectrum. (Hint: what properties does this operator have?)

EDIT: Discussed conclusion briefly in comments

Answer 2

The point spectrum $\sigma_p(T)$ of $T$ is precisely the set of eigenvalues of $T$; if

$Tf = \lambda f, \tag 1$

then

$\displaystyle \int_0^x f(s) \; ds = Tf = \lambda f(x); \tag 2$

since

$f(x) \in C[0, 1], \tag 3$

any such $f(x)$ satisfying (1) with $\lambda \ne 0$ is continuously differentiable; thus for such $\lambda$ we may write

$f(x) = \lambda f'(x); \tag 4$

the unique solution to this ordinary differential equation is well-known to be

$f(x) = f(0)e^{x/\lambda}, \; x \in [0, 1]; \tag 5$

however, when (5) is substituted into (2) we obtain

$\lambda f(x) = \displaystyle \int_0^x f(0)e^{s/\lambda} \; ds = f(0) \lambda(e^{x/\lambda} - 1)$ $= f(0)\lambda e^{x/\lambda} - f(0)\lambda = \lambda f(x) - \lambda f(0); \tag 6$

with $\lambda \ne 0$ this yields

$\lambda f(0) = 0 \Longrightarrow f(0) = 0 \Longrightarrow f(x) = 0; \tag 7$

but $f(x) = 0$ is by definition not admissible as an eigenvector/eigenfunction; thus $\sigma_p(T)$ consists of at most $0$, which is certainly not an admissible eigenvalue of $T$ either, for then (2) becomes;

$\displaystyle \int_0^x f(s) \; ds = 0, \tag 8$

which differentiated implies

$f(x) = 0, \; \forall x \in [0, 1], \tag 9$

which is not an admissible eigenvector. Therefore,

$\sigma_p(T) = \emptyset. \tag{10}$