Let $K$ be a psd kernel on an abstract space $X$ and let $H_K$ be the induced Reproducing Kernel Hilbert Space (RKHS). Let $P$ be a probability measure on $X$ such that $H_K \subseteq L^2(P_X)$ and suppose the self-adjoint kernel integral operator $T_K:L^2(P) \to H_K$ defined by $(T_Kf)(x) := \int_X K(x,x') f(x')\,dP$ is compact.
Question 1. Under what conditions on $K$ and $P$ does there exist a constant $0 < C < \infty$ such that $$ \|f\|_{H_K} \le C\|f\|_{L^2(P)}, $$ for all $f \in H_K$ ?
In particular, I'm interested in the case where $X = \mathbb R^d$ and $P$ is the standard Gaussian measure on $\mathbb R^d$.
Observation
Note that one always has $\|f\|_{H_K} = \|T_K^{1/2} f\|_{L^2(P)} \le \sqrt{\|T_K\|_{op}}\|f\|_{H_K}$ for any $f \in H_K$. Indeed,
It is well-known from Mercer's Theorem that if $T_K f := \sum_n \lambda_n \langle f,\phi_n\rangle_{L^2(P)}$ is the spectral decomposition of $T_K$ (where the $\phi_n$'s are an orthonormal basis for $L^2(P)$), then the inner-product in $H_K$ is given by $$ \langle f,g\rangle_{H_K} = \sum_n \frac{\langle f,\phi_n\rangle_{L^2(P)}\langle g,\phi_n\rangle_{L^2(P)}}{\lambda_n}, $$ and moreover $H_K$ is given by $$ H_K = \left\{f \in L^2(P) \mid \sum_n \frac{\langle f,\phi_n\rangle_{L^2(P)}^2}{\lambda_n} < \infty\right\}. $$
One then deduces that $$ \begin{split} \|T_K^{1/2} f\|_{H_k}^2 &= \langle T_K^{1/2}f,T_K^{1/2} f\rangle_{H_k} = \langle f,T_K f\rangle_{H_k}\\ & = \sum_n \dfrac{\langle f,\phi_n\rangle_{L^2(P)}\langle T_K f,\phi_n\rangle_{L^2(P)}}{\lambda_n}\\ & = \sum_n \dfrac{\langle f,\phi_n\rangle_{L^2(P)}\langle f,T_K\phi_n\rangle_{L^2(P)}}{\lambda_n}\\ &= \sum_n \langle f,\phi_n\rangle_{L^2(P)}^2 = \|f\|_{L^2(P)}^2. \end{split} $$
Thus, $\|f\|_{L^2(P)} = \|T_K^{1/2}f\|_{H_K} \le \sqrt{\|T_K\|_{op}}\|f\|_{H_K}$.
