# Functions in a Reproducing Kernel Hilbert Space are Lipschitz continuous

I would like to show that all the functions in a Reproducing Kernel Hilbert Space (RKHS) are Lipschitz continuous. So that, I take two points in the domain $\vec{x}_{1} ,\vec{x}_{2} \in X$ then from the Riez Representation Theorem it follows that,

$$\begin{array}{l} {\left|f\left(\vec{x}_{1} \right)-f\left(\vec{x}_{2} \right)\right|=\left|\left\langle f,K_{\vec{x}_{1} } \right\rangle _{H_{K} } -\left\langle f,K_{\vec{x}_{2} } \right\rangle _{H_{K} } \right|} \\ {\quad \quad \quad \quad \quad \quad \; =\left|\left\langle f,K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\rangle _{H_{K} } \right|} \\ {\quad \quad \quad \quad \quad \quad \; \le \left\| f\right\| _{H_{K} } \left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\| _{H_{K} } } \end{array}$$

Here $H_{K}$ is a RKHS and the inequality comes from Cauchy-Schwarz inequality. In a document online, it is that $\left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\|=(K_{\vec{x}_{1} }-K_{\vec{x}_{2} })^{2}$. But I can not show it. I am pretty sure that it equals to difference of their kernelized positions, i.e., $\left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2}} \right\|=d(K_{\vec{x}_{1}},K_{\vec{x}_{2}})=\sqrt{\left\langle(K_{\vec{x}_{1}}-K_{\vec{x}_{2}}),(K_{\vec{x}_{1}}-K_{\vec{x}_{2}})\right\rangle}$

Much obliged for your help in advance.

Pag.28 of these slides may be useful to you.

• Please summarize the Ideas on the Slide. It's always better to present the information directly on the questions/answer environment. – bertozzijr Jul 27 at 14:02