I'm trying to solve the following exercise:
Let $f\in\mathcal{C}([0,1])$ and let $T$ an operator such that $Tf(x)=\int_0^1(x-t)f(t)dt$. I have proved that $T$ is a bounded linear operator and, by means of Ascoli-Arzelà theorem, that it is a compact operator.
Now I need to find its kernel, its rank (showing a basis) and its spectrum. I'm quite stucked, without an idea which could make me start.
Thank you for any suggestion!
As Chris Eagle said, $\ker T=\ker L_1\cap \ker L_2$, where $L_1$ and $L_2$ are two linear continuous functionals on $\mathcal C([0,1])$ defined by $L_1(f)=\int_0^1f(t)dt$ and $L_2(f)=\int_0^1tf(t)dt$. We have $$\operatorname{rank}(T)=\left\{x\mapsto ax+b,a,b\in\mathbb C\right\}.$$ Indeed, the formula $T(f)(x)=x\int_0^1f(t)dt-\int_0^1tf(t)dt$ show that each element of the range should have this form. Since $T(2)(x)=2x-1$ and $T(6X)(x)=3x-2$, we can get all these functions.
For the spectrum, take $\lambda\neq 0$. An eigenvector should have the form $f(t)=at+b$ with $a,b\in\mathbb C$. We get the system $$\left\{\begin{array}{cc} \left(\lambda-\frac 12\right)a-b&=0\\\ \frac a3+\left(\lambda+\frac 12\right)b&=0, \end{array} \right. $$ which have a non-trivial solution if and only if $\lambda^2-\frac 14+\frac 13=0$ i.e. $\lambda^2+\frac 1{12}=0$, so $\lambda=\pm\frac i{2\sqrt 3}$. We can check that for these $\lambda$, we have indeed an eigenvector, namely $f(x)=2\sqrt 3x-\sqrt 3\pm i$ for example. Since $T$ is compact in a infinite dimensional vector space, the spectrum is $\{0\}\cup \{\mbox{eigenvalues}\}$, so $\sigma(T)=\left\{0,\frac i{2\sqrt 3},-\frac i{2\sqrt 3}\right\}$.
