# How do I prove that Euclidean Distance Function is a valid kernel?

Say a Euclidean distance function $d$ is given as: $d(x, y) = \Sigma(x_i - y_i)^2$.

How do I prove it is a valid kernel?

I know this: $d(x, y) = \Sigma(x_i - y_i)^2 = \Sigma(x_i^2 + y_i^2 - 2x_iy_i) = \langle x, x\rangle + \langle y, y\rangle - 2\langle x, y\rangle$

However, how do i prove that the difference of two kernels in this case is a valid kernel?

• what do you mean by "valid kernel"? – utobi Dec 5 '16 at 8:04
• that the function is symmetric and positive semidefinite – lucifer1190 Dec 5 '16 at 8:52

Your (squared Euclidean) distance function $d(x,y) = \sum(x_i - y_i)^2$ is quadratic and therefore $d(x,y) \geq 0 \,\, \forall x,y \in \mathbb{R}$, i.e. your kernel is positive definite. The second necessary condition for a valid kernel is symmetry, $d(x,y) = d(y,x)$, which is also fulfilled in your case.