Given a self-adjoint finite rank integral operator P on $L_2[0,1]$, it has the eigen-decomposition $P=\sum_{i=1}^k\lambda_i \langle u_i,\cdot\rangle u_i$ where $u_i$ are eigenfunctions and $\lambda_i$ are eigenvalues for P.

Let $p(x,y)$ be the kernel for $P$, i.e. $Pf(x)=\int_0^1 p(x,y)f(y)dy$, what can we say about $p(x,y)$? Is it always true that there are finite number of $f_k$ which are orthogonal and we can write $p(x,y)=\sum_{i,i'} \langle f_i,Pf_i'\rangle f_i(x)f_i'(x)$ ? If such expression exists, what are the relationship between $f_i$ and the eigenfunctions of P, namely $u_j$ ?

We can assume $0<p(x,y)<C$ for some constant C.



1 Answer 1


From the fact that $P$ is selfadjoint, you can deduce that $u_1,\ldots,u_k$ are orthonormal: for $j\ne m$, $$ \lambda_j\langle u_j,u_m\rangle=\langle Pu_j,u_m\rangle=\langle u_j,Pu_m\rangle=\lambda_m\langle u_j,e_m\rangle. $$ So $\langle u_j,u_m\rangle=0$. And then $Pu_j=\lambda_j\|u_j\|^2\,u_j$, so $\|u_j\|=1$.

Then you can take $$ p(x,y)=\sum_{j=1}^k\lambda_j u_j(x)u_j(y). $$ You have $$ \int_0^1\sum_{j=1}^k\lambda_j u_j(x)u_j(y)\,f(y)\,dy=\sum_{j=1}^k\lambda_j u_j(x)\int_0^1 u_j(y)\,f(y)\,dy=\sum_{j=1}^k\lambda_j \langle u_j,f\rangle\,u_j(x)=(Pf)(x). $$


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