# problem over positive definite kernel.

I'm having the following problem for which I can't find a solution.

Let's consider two kernel, $$K(x,y)$$ and $$K'(x,y)$$, function from $$X^2 \rightarrow \mathbb{R}$$ with $$X$$ an arbitrary set. I have the following property, $$\forall (x,y), K(x,y) > K'(x,y)$$ and kernel $$K'$$ is positive definite, meaning that the following property holds :

$$\sum_{i=1}^n \sum_{j=1}^n c_i c_j K'(x_i,x_j) \geq 0$$

for $$c_1,..,c_n \in \mathbb{R}$$. Is it possible to show that in this case the kernel $$K(x,y)$$ is positive definite ? My intuition is that we can't infer anything on $$K$$ just from knowing $$K(x,y) > K'(x,y)$$.

It's not exactly clear what you are assuming about $$K$$, but it is not true that if a symmetric matrix (such as $$A=\pmatrix{10&20\\20&10}$$) is element-wise greater than a positive definite matrix (such as $$B=\pmatrix{1&0\\0&1}$$), that the first matrix must necessarily be positive definite. For example, $$c=(1,-1)^T$$ has the property that $$c^TAC=-20$$ but $$c^TBc=2$$.
If your assumption on $$K$$ is that its elements form a semi-positive definite matrix, and your question is, does the element-wise inequality $$K>K'$$ imply $$K$$ is strictly psd if $$K'$$ is, the answer is no. $$A=\pmatrix{10&10\\10&10}$$ gives an example of a semi-definite matrix which is strictly element-wise greater than a strictly positive definite matrix.