# RKHS norm of Lipschitz functions

Given a set $$\mathcal{X}$$ and RKHS $$\mathcal{H}$$ of functions on $$\mathcal{X}$$, we can recover a (pseudo)metric on $$\mathcal{X}$$ by $$d(x,y)=||\phi_x-\phi_y||_{\mathcal{H}}$$, where $$\phi_x=k(x,\cdot)$$.

It is straightforward to see that any function $$f \in \mathcal{H}$$ which has RKHS norm less than $$L$$ is Lipschitz (with respect to our metric above) with constant $$L$$:

$$|f(x)-f(y)|=|\langle f,\phi_x-\phi_y\rangle|\leq ||f||_{\mathcal{H}}d(x,y),$$ for any $$x,y \in \mathcal{X}$$.

I am very interested in the following question: if we have a function $$f\in \mathcal{H}$$ which is Lipschitz with constant $$L$$, is there anything we can say about it's norm $$||f||_{\mathcal{H}}$$?

• by "anything we can say", do you mean bounding it from above? If yes, then you want your Hilbert space to be isomorphic to a subspace of a Lipschitz space. I would expect a subspace of the Lipschitz space isomorphic to a Hilbert space necessarily be finitely dimensional, but I am not sure. – erz Jul 4 at 9:27
• What happens when $\mathcal{X}=\mathbb{Z}$ and $\mathcal{H}=l^2$? – DCM Jul 4 at 11:39
• @erz yes, bounding it above was precisely what I was looking for – Tyler6 Jul 4 at 13:09

I think that in general $$L(.)$$ and $$\Vert.\Vert_\mathcal{H}$$ measure quite different things.
Writing $$L(f)$$ for
$$\inf\{ M>0:|f(x)-f(x')| \leq Md(x,y) \;\forall \;x,x'\in \mathcal{X}\}$$
let $$\mathcal{X}=\mathbb{Z}$$ and $$\mathcal{H}=l^2$$. Then (unless I've made an embarrassing mistake...) setting $$f_n=1_{[-n,n]}$$ gives you a sequence in $$\mathcal{H}$$ with $$L(f_n)=1/2$$ and $$\Vert f_n\Vert_\mathcal{H}=2n$$.
• Hmm, correct me if I’m wrong but I thought $l^2$ wasn’t an RKHS? Either way it’s an interesting counterexample – Tyler6 Jul 4 at 13:06
• I think another example is $\mathcal{H} = H^1_0((0,1])$, i.e. the absolutely continuous functions $f : [0,1] \to \mathbb{R}$ with $f' \in L^2([0,1])$, under the norm $\|f\|_{\mathcal{H}}^2 = \int_0^1 |f'|^2$. Then the "Lipschitz constant" is the Holder norm of exponent 1/2, $L = \sup |f(s)-f(t)|/\sqrt{|s-t|}$. So now let $f_n$ approach $f(x) = \sqrt{x}$ which has infinite $\mathcal{H}$-norm. – Nate Eldredge Jul 4 at 17:19