$K(x,y)\in L^{\infty}(R^n\times R^n, m\times m)$, $K(x,y)=K(y,x)$, so $K(x,y)=\sum_{k=1}^{\infty}\lambda_k \phi_k(x)\phi_k(y)$, are $\phi_k$ bounded?

Question

Consider a symmetric function $$ K(x,y):R^n \times R^n \to R $$ satisfying $K(x,y)=K(y,x)$ and $$ \int_{R^n\times R^n} K^2(x,y)dm(x) dm(y) <\infty. $$ Let $m$ be a probability measure on $R^n$.

Thanks to Iosif Pinelis's anwer If $f(x_1,x_2)=f(x_2,x_1)$, $f(x_1,x_2)=\sum_k \lambda_k f_k(x_1)f_k(x_2)$? $$ Tf(x)=\int_{R^n} K(x,y)f(y)dm(y) $$ is a bounded, self-adjoint operater from $f\in L^2(R^n, m)$ to $L^2(R^n, m)$. Then there exist $\lambda_k$ and orthonormal basis $\phi_k \in L^2(R^n,m)$ such that $$ \int_{x\in R^n}\phi_k(x)\phi_l(x)dm=\delta_{kl}, $$ $T\phi_k=\lambda_k \phi_k$, $f=\sum_{k=1}^{\infty} a_k \phi_k$ (convergence w.r.t $L^2(R^n,m)$）and $$ K(x,y)=\sum_{k=1}^{\infty} \lambda_k \phi_k(x)\phi_k(y) $$ (Convergence w.r.t. $L^2(R^n\times R^n, m\times m)$).

If we assume in addition that $K(x,y)$ is bounded on $R^n\times R^n$, can we get that $\phi_k $ are bounded on $R^n\times R^n$ or $\phi_k \in L^p(R^n,m)$ for some $p>2$? If it's wrong, can you give a counter example?

Answer 1

Because $m(R^n) = 1$, we have $L^\infty(R^n,m) \subset L^2(R^n,m) \subset L^1(R^n,m)$. So if an eigenvector $\phi_k$ is in $L^2(R^n)$, it is also in $L^1(R^n,m)$. By the usual inequality $\|Tf\|_{\infty} \le \|K(-,-)\|_\infty \|f\|_1$, so that $\phi_k \in L^2(R^n,m) \subset L^1(R^n,m)$ implies that $\phi_k = \frac{1}{\lambda_k} T\phi_k \in L^\infty(R^n,m)$ as long as $\lambda_k \ne 0$.

I'm not sure if your question extends to the possibility that some of your $\lambda_k=0$. The the question reduces to the following: does every closed subspace of $L^2(R^n,m)$ whose orthogonal complement has a Hilbert basis where each element belongs to $L^\infty(R^n,m)$ itself have a Hilbert basis of the same type? Maybe the answer to that is obvious, but it escapes me at the moment.