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.
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 $\int_{\mathcal X }|k(x,x)|^2\leqslant 1$.