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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 $||f||_{\mathcal{H}}$?

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    $\begingroup$ 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. $\endgroup$
    – erz
    Jul 4, 2020 at 9:27
  • $\begingroup$ What happens when $\mathcal{X}=\mathbb{Z}$ and $\mathcal{H}=l^2$? $\endgroup$
    – DCM
    Jul 4, 2020 at 11:39
  • $\begingroup$ @erz yes, bounding it above was precisely what I was looking for $\endgroup$
    – Tyler6
    Jul 4, 2020 at 13:09

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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 $\Vert f_n\Vert_\mathcal{H}=2n$.

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  • $\begingroup$ Hmm, correct me if I’m wrong but I thought $l^2$ wasn’t an RKHS? Either way it’s an interesting counterexample $\endgroup$
    – Tyler6
    Jul 4, 2020 at 13:06
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    $\begingroup$ What's your definition of a RKHS? I think the usual one is just "a Hilbert space of functions on a set whose evaluation functionals are norm continuous", no? $\endgroup$
    – DCM
    Jul 4, 2020 at 13:37
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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Jul 4, 2020 at 17:19

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