# $d(x,y)$ in terms of reproducing kernels for a H separating points

I'm working through an introduction to the Theory of Reproducing Kernel Hilbert Spaces by V.I. Paulsen and M. Raghupathi.

Exercise 1.1 states, that let $$\mathcal{H}$$ be the reproducing kernel hilbert space of $$X$$, $$d(x,y,):=\sup\{|f(x)-f(y)|: f\in\mathcal{H}, \|f\|\le 1\}$$ is a metric if and only if $$\mathcal{H}$$ separates points.

I was able to prove this statement. However, they go on and say give a formula for $$d(x,y)$$ in terms of the reproducing kernel. What I achieved so far

$$|f(x)-f(y)|=\|\langle f,k_x\rangle-\langle f,k_y\rangle\| \le\|f\|\|k_x-k_y\|\le\|k_x-k_y\|$$

i.e. $$d(x,y)\le\|k_x-k_y\|$$. Now clearly would be nice to show that this bonudary is sharp and I guess if this is the case we would need the separation property. Can someone give me a hint, please?

As you note you have that $$|f(x)-f(y)|=| \langle f, k_x-k_y\rangle| ≤ \|k_x-k_y\|$$
Giving that $$\sup_{\|f\|≤1}|f(x)-f(y)|≤\|k_x-k_y\|$$. Now look at $$f= \frac{k_x-k_y}{\|k_x-k_y\|}$$, which is well defined (since $$\mathcal H$$ separates points) and norm one. You have that
$$\langle\frac{k_x-k_y}{\|k_x-k_y\|}, k_x-k_y\rangle = \frac{\|k_x-k_y\|^2}{\|k_x-k_y\|} =\|k_x-k_y\|$$ hence $$\sup_{\|f\|≤1} |f(x)-f(y)|≥\|k_x-k_y\|$$.