My question extends this one: 1. In Hermann Koenig's book: Eigenvalue distribution of compact operators (page 145), the Mercer's theorem is stated with an extra claim. Given a Mercer's kernel $k$, the Hilbert-Schmidt integral operator is defined as

$T_k: (T_k f)(x) = \int_y k(x,y) f(y) dy$

Then the book claims that the eigenfunctions for $T_k$ are uniformly bounded in the $L_\infty$ norm: $\sup_{n} \|f_n\|_\infty < \infty$. Here eigenfunctions $f_n$ have unit $L_2$ norm.

The answers in the above link have constructed a counter-example. However I'm wondering if there is any natural condition we can put on the kernel in order to guarantee the uniform boundedness. For example, it surely holds true for the standard Gaussian RBF kernel. But how about the kernel $k(x,y) = \frac{1}{2-x^\top y}$ over the unit ball in $R^n$? It's sometimes called inverse kernel, but I don't want to confuse with the "inverse of the kernel".

My question extends this one: 1. In Hermann Koenig's book: Eigenvalue distribution of compact operators (page 145), the Mercer's theorem is stated with an extra claim. Given a Mercer's kernel $k$, the Hilbert-Schmidt integral operator is defined as

$T_k: (T_k f)(x) = \int_y k(x,y) f(y) dy$

Then the book claims that the eigenfunctions for $T_k$ are uniformly bounded in the $L_\infty$ norm: $\sup_{n} \|f_n\|_\infty < \infty$. Here eigenfunctions $f_n$ have unit $L_2$ norm. See a scanned copy of the page here: 2.

The answers in the above link have constructed a counter-example. However I'm wondering if there is any natural condition we can put on the kernel in order to guarantee the uniform boundedness. For example, it surely holds true for the standard Gaussian RBF kernel. But how about the kernel $k(x,y) = \frac{1}{2-x^\top y}$ over the unit ball in $R^n$? It's sometimes called inverse kernel, but I don't want to confuse with the "inverse of the kernel".

My question extends this one: 1. In Hermann Koenig's book: Eigenvalue distribution of compact operators (page 145), the Mercer's theorem is stated with an unusual claim. Given a Mercer's kernel $k$, the Hilbert-Schmidt operator is defined as

$T_k: (T_k f)(x) = \int_y k(x,y) f(y) dy$

Then the book claims that the eigenfunctions for $T_k$ are uniformly bounded in the $L_\infty$ norm: $\sup_{n} \|f_n\|_\infty < \infty$. Here eigenfunctions $f_n$ have unit $L_2$ norm.

The answers in the above link have constructed a counter-example. However I'm wondering if there is any natural condition we can put on the kernel in order to guarantee the uniform boundedness. For example, it surely holds true for the standard Gaussian RBF kernel. But how about the inverse kernel $k(x,y) = \frac{1}{2-x^\top y}$ over the unit ball in $R^n$?

My question extends this one: 1. In Hermann Koenig's book: Eigenvalue distribution of compact operators (page 145), the Mercer's theorem is stated with an extra claim. Given a Mercer's kernel $k$, the Hilbert-Schmidt integral operator is defined as

$T_k: (T_k f)(x) = \int_y k(x,y) f(y) dy$

Then the book claims that the eigenfunctions for $T_k$ are uniformly bounded in the $L_\infty$ norm: $\sup_{n} \|f_n\|_\infty < \infty$. Here eigenfunctions $f_n$ have unit $L_2$ norm.

The answers in the above link have constructed a counter-example. However I'm wondering if there is any natural condition we can put on the kernel in order to guarantee the uniform boundedness. For example, it surely holds true for the standard Gaussian RBF kernel. But how about the kernel $k(x,y) = \frac{1}{2-x^\top y}$ over the unit ball in $R^n$? It's sometimes called inverse kernel, but I don't want to confuse with the "inverse of the kernel".