According to this paper (just right below the Theorem 3 and above the section 3)

Reproducing Kernel Hilbert Space(RKHS) $\mathcal{H}$ on $\mathcal{X}$ is a Hilbert space of functions from $\mathcal{X}$ to $\mathbb{R}$. $\mathcal{H}$ is an RKHS if and only if there exists a $k(x,x'):\mathcal{X}\times\mathcal{X}\to\mathbb{R}$ such that $\forall x\in\mathcal{X},\,k(x,\cdot)\in\mathcal{H}$, and $\forall f\in\mathcal{H},\, \langle f,k(x,\cdot)\rangle_{\mathcal{H}}=f(x)$. If such a $k$ exist, it is unique and it is a PD kernel. A function $f\in\mathcal{H}$ if and only if $\|f\|^2_{\mathcal{H}}=\langle f,f\rangle_{\mathcal{H}}<\infty$, and its $L_2$ norm is dominated by RKHS norm $\|f\|_{L_2}\le \|f\|_{\mathcal{H}}$.

How the last claim can be done ?

I thought this would be fundamentally mathematical, and the claim and related definitions are independent of this specific paper, thus it might be understood especially by people who had some background knowledge. I want to know how to prove this claim, and I am a beginner and even do not know how to describe $\|f\|_{L_2} \leq \|f\|_{\mathcal{H}} $ explicitly .

Thanks for all your help.

  • $\begingroup$ Begin by stating the definition of $\mathcal H$. $\endgroup$
    – user147263
    Dec 26, 2014 at 3:30
  • $\begingroup$ I don't really understand what you're asking. By definition, a Hilbert space is a (complete) inner product space. So if $f\in\mathcal H$, then necessarily $\langle f,f\rangle$ is a complex number (with imaginary part zero by conjugate symmetry but that isn't important here); there is nothing to prove there. As for the second part, I imagine it would depend on which measure space with respect to which you're taking the $L^2$ norm... $\endgroup$
    – Math1000
    Dec 26, 2014 at 3:31
  • $\begingroup$ Sorry, I thought it would be a fundamentally mathematical question. The question is rxtracted from the paper cs.cmu.edu/~ninamf/papers/scalable-kernels.pdf , just right below the Theorem 3 and above the section 3 of the paper. $\endgroup$
    – olivia
    Dec 26, 2014 at 9:09

1 Answer 1


Use the definition with $f\colon x'\mapsto k(x,x')$: we get $$k(x,x) =\lVert k(x,\cdot)\rVert_{\mathcal H} ^2.$$ By Cauchy-Schwarz inequality, we get $$|f(x)|^2\leqslant|k(x,x)|^2\lVert f\rVert_\mathcal H^2,$$ hence $\lVert f\rVert_{L^2}\leqslant \lVert f\rVert_\mathcal H $ provided that $\int_{\mathcal X }|k(x,x)|^2\leqslant 1$.

  • $\begingroup$ Should it be $|k(x,x)|^2$ as written or $|k(x,x)|$? $\endgroup$
    – wij
    Oct 6, 2021 at 21:29

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