Eigenvalues and eigenfunctions of an integral operator Let $T$ be an integral operator with kernel $K(x,y)=e^{|x-y|}$ on $L^2(-1,1)$. How can we find the eigenfunctions and eigenvalues of $T$?
Even though I am not sure whether the following arguments are correct, here is what I have done :
Say $Tf(x) = \displaystyle\int_{-1}^1 e^{|x-y|}f(y)dy.$ When $x-y \gt 0$ we have 
 $Tf(x) = e^x \displaystyle\int_{-1}^1 e^{-y} f(y)dy. $ Suppose that we have $Tf(x) = \lambda f(x)$ for some scalar $ \lambda$.
Evaluating the integral, we have $Tf(x) = e^x (e-e^{-1}) = \lambda f(x)$. 
Now, taking derivatives of both sides, $e^x(e-e^{-1})  = \lambda f'(x)$. Thus, we have $\lambda f(x) = \lambda f'(x) = \dots$. This seems like an ODE problem but we need some boundary coundtion and here is where I am stuck.
 A: This is not an answer : Just some calculations for the asker I couldn't place into a comment.
Clearly, considering the two cases 


*

*$x-y>0 \ \iff \ y<x$

*$x-y<0 \ \iff \ y>x$
the integral can be split in the following way :
$$(Tf)(x)=e^{x} \underbrace{\int_{y=-1}^{y=x} e^{-y}f(y)dy}_{f_1(x)} \ + \ e^{-x} \underbrace{\int_{y=x}^{y=1} e^{y}f(y)dy}_{f_2(x)}\tag{1}$$
In particular :
$$(f_1)'(x))=e^{-x}f(x) \ \ \text{and} \ \ (f_2)'(x))=-e^{x}f(x).\tag{2}$$
We have 
$$(Tf)'(x) \ = \ e^{x}(f_1(x)+f'_1(x))+e^{-x}(-f_2(x)+f'_2(x))\tag{3}$$
Taking (2) into account, one gets :
$$(Tf)'(x)=e^{x}(f_1(x)+e^{-x}f(x))+e^{-x}(f_2(x)-e^{x}f(x))\tag{4}$$
i.e., by cancellation :
$$(Tf)'(x)=e^{x}f_1(x)-e^{-x}f_2(x)\tag{5}$$
Differentiating (5) gives :
$$(Tf)''(x)=e^{x}(f_1(x)+f'_1(x))-e^{-x}(-f_2(x)+f'_2(x))\tag{6}$$
i.e.,
$$(Tf)''(x)=e^{x}(f_1(x)+e^{-x}f(x) )-e^{-x}(-f_2(x)-e^{x}f(x))\tag{7}$$
$$(Tf)''(x)=\underbrace{e^{x}f_1(x)+e^{-x}f_2(x)}_{Tf(x)}+2 f(x)\tag{8}$$
Knowing that $(Tf)(x)=\lambda f(x)$ (eigenvalue/eigenfunction), can you take it from here (classical second order differential equation) ?
