Conditions for transformations of kernels to be kernels

Question

We are given the results that if $(k_n)_{n\in\mathbb{N}}$ are kernels (positive definite symmetric functions) then $k_1+k_2$, $k_1 k_2$ and $\lim_{n\to\infty}k_n$ (if it exists) are kernels, then we can deduce that $\sum_{n\in\mathbb{N}} k_n$ and $k_1^m$ are kernels too in case they exists.

Is it then true that if $\hat{k}$ is a kernel and $f$ is a function that can be expressed as a power series with positive coefficients $a_n≥0$: $f(t)=\sum_{n\in\mathbb{N}} a_n (t-t_0)^n$ then $f\circ \hat{k}$ is also a kernel because:

$$(f\circ\hat{k})(x,y)=\sum_{n\in\mathbb{N}} a_n (\hat{k}(x,y)-t_0)^n$$

is the limit of a sum of a linear combination with positive coefficientes of kernels. This would be useful to prove a lot of functions to be kernels. I cannot seem to come with a reason for this to be wrong, but I haven't seen the result anywhere so I'm not sure.

Answer 1

If you relax "positive definite" to "positive semi-definite", then it is not hard to show that $k_f:=f\circ\hat k$ is indeed a kernel, provided that $t_0\le 0$. Here's a sketch of proof :

• for any constant $c\ge 0$, the map $(x,y)\mapsto c$ is a kernel,
• if $k_1$ and $k_2$ are kernels, and $\lambda \ge 0$ a constant, then $k_1 + \lambda k_2$ is also a kernel. This follows from the first bullet-point together with your assumptions about sum and product of kernels.
• if $k$ is a kernel, then so is $k^n :(x,y)\mapsto k(x,y)^n$ for any natural $n$ (by induction),
• from the above it follows that if $(a_n)_{n\ge 1}$ is a sequence of non-negative real numbers and $t_0\le 0$, then for any $N\ge 1$ $$ k_f^{(N)} :(x,y)\mapsto \sum_{n=1}^N a_n\left(\hat k(x,y)-t_0\right)^n $$ is also a kernel.
• The only thing left now is to check that the pointwise limit $k_f$ is also positive definite. This follows because, for all $m\ge 1$ and $(x_1,u_1),\ldots,(x_m,u_m)\in \mathcal X\times \mathbb R$, we have $$\sum_{i=1}^m u_i k_f(x_i, x_j) u_j = \lim_{N\to\infty} \sum_{i=1}^m u_i k_f^{(N)}(x_i, x_j) u_j\ge 0.$$

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However if you require that $k_f$ is positive definite, then the answer can be no : in general we can't expect $k_f$ to be positive definite. You can have a look at the counterexamples provided in these two answers :

• Gaussian kernels for arbitrary metric spaces
• Is the exponential of $−d$ a positive definite kernel for a general metric space $(X,d)$ ?