Eigenfunctions and eigenvalues of an operator defined by a certain integral Let $k :[0,1]^2 \rightarrow \bf{R} $ be a kernel function definded by
$ k(x,y)= (1- max(x,y))^2 .$ Now, let $ L $ be a linear operator defined on $ L^2 [0,1] $  by $$ Lf(x):=\int_0^1 k(x,y)f(y)dy$$. Can we find eigenfunctions and the associated eigenvalues? (I'm looking for eigenfunctions which forms an orthonormal basis. cf. Hilbert-Schmidt Theorem)
 A: The inverse operator is $L^{-1} = 1/2 \cdot (x-1)^{-1} d^2/dx^2 - 1/2 \cdot (x-1)^{-2} d/dx$. Its eigenfunctions are derivatives of Airy functions, $Ai^{\prime } ((2\lambda )^{1/3} (x-1))$, $Bi^{\prime } ((2\lambda )^{1/3} (x-1))$, where $\lambda $ are the corresponding eigenvalues. Without further specification of the space of functions $f$, i.e., boundary conditions, no further restriction is placed on $\lambda $.
EDIT: Spurred by Christian Remling's comments, the boundary conditions and the resulting quantization of eigenvalues can be further specified: Since $k(1,y)=0$, solutions must vanish at $x=1$, implying that we have to combine the Airy function derivatives as $\sqrt{3} Ai^{\prime } ((2\lambda )^{1/3} (x-1)) + Bi^{\prime } ((2\lambda )^{1/3} (x-1))$. The allowed eigenvalues $\lambda $ are then determined by the condition that $-(2\lambda )^{1/3} $ must correspond to one of the extrema of the function $\sqrt{3} Ai^{\prime } + Bi^{\prime } $.
These eigenfunctions of $L^{-1} $ are also the eigenfunctions of $L$, with eigenvalues $1/\lambda $.
