Let $T$ be an operator in $L^2[0,1]$, so that $T$ is an integral operator:
$$Tf(x)=\int_0^1k(x,y)f(y)dy$$
Where $k(x,y)=1-x, y\le x$ and $k(x,y)=1-y, y\ge x$.
Show that $T$ has a spectral representation, and find it.
Since $T$ is an integral operator, we can show that it is compact. Since $k(x,y)=k(y,x)$ and is over $R$, we have that $T$ is self-adjoint. Thus by the spectral representation theorem, we have that $T$ has one.
Now, in order to find the spectral representation, we first need to find the eigenvalue and eigenvectors.
Assuming $Tf(x)=\lambda f(x)$, we have: $$\int_0^x (1-x)f(y)dy + \int_x^1 (1-y)f(y)dy = \lambda f(x)$$
Without lose of generality, we may assume the result is true for every $x$, and use derivatives. Simplifying the results, we have: $$f(x)=-\lambda f''(x)$$
Now this is where it gets messy. How do I get from here to the spectral representation? This seems complicated, over the top. Is there a better way to approach the representation itself? Any guidance\hints will be appreciated!
Hint: from the integral equation you can also get boundary conditions $\lambda f(1)=0$ and $\lambda f'(0)=0$. The ODE with this conditions has the trivial solution (which is not an eigenvector), unless $\lambda\ne 0$ is a very special number for the equation + the interval.