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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?

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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 $\sup_{\|f\|≤1} |f(x)-f(y)|≥\|k_x-k_y\|$.

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  • $\begingroup$ ah...that was simple than thought. thanks for the answer. I will award the bounty in 20hours (it's locked for 24hours) $\endgroup$
    – math
    Jul 22, 2020 at 12:13

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