Does $c$ in the embedding inequality $\|P^\kappa_t \|_{L^p \to L^{p'}} \le c t^{-\frac{d(p'-p)}{2pp'}}$ depend on $\kappa$?

Question

For any $\kappa>0$, we consider the Gaussian heat kernel $$ p^\kappa_t (x) := (\kappa \pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{\kappa t}}, \quad t>0, x \in {\mathbb R}^d. $$

Let $L^0 := L^0 (\mathbb R^d)$ be the space of real-valued measurable functions on $\mathbb R^d$. Let $L^0_+ := L^0_{+} (\mathbb R^d)$ the subspace of $L^0$ consisting of non-negative functions. Let $L^0_b := L^0_b (\mathbb R^d)$ the subspace of $L^0$ consisting of bounded functions. For $f \in L^0_b \cup L^0_+$, let $$ P_t^\kappa f (x) := \int_{\mathbb R^d} p^\kappa_t (x-y) f (y) \, \mathrm d y, \quad t >0, x \in \mathbb R^d. $$

It is stated in the paper Singular density dependent stochastic differential equations that

It is well-known that for some constant $c>0$, $$ \|P^\kappa_t \|_{L^p \to L^{p'}} := \sup_{\|f\|_{L^p} \le 1} \|P^\kappa_t f\|_{L^{p'}} \le c t^{-\frac{d(p'-p)}{2pp'}}, \quad t>0, 1 \le p \le p' \le \infty. $$

I would like to ask if the constant $c$ depends on $\kappa$. Any reference is appreciated.

Thank you so much for your elaboration!

Answer 1

$\newcommand\ka\kappa$Yes, the best constant $c$ depends on $\ka$ if $p<p'<\infty$.

Indeed, the inequality in question can be rewritten as $$t^Q\sup_{\|f\|_{L^p} \le 1} \|P^\ka_t f\|_{L^{p'}} \le c \tag{1}\label{1}$$ if $t>0$ and $1 \le p \le p' \le \infty$, where $Q:=\frac{d(p'-p)}{2pp'}\in[0,\infty]$. So, the best constant $c$ in inequality \eqref{1} is $$c^\ka:=\sup_{t>0}t^Q\sup_{\|f\|_{L^p} \le 1} \|P^\ka_t f\|_{L^{p'}}.$$

Note that $p^\ka_t=p^1_{\ka t}$ and hence $P^\ka_t=P^1_{\ka t}$. So, $$c^\ka=\sup_{t>0}t^Q\sup_{\|f\|_{L^p} \le 1} \|P^1_{\ka t} f\|_{L^{p'}} =\ka^{-Q}\sup_{u>0}u^Q\sup_{\|f\|_{L^p} \le 1} \|P^1_u f\|_{L^{p'}} =\ka^{-Q}c^1,$$ which depends on $\ka>0$ if $Q\ne0$, since $c^1\in(0,\infty)$ (the latter fact follows immediately from Young's convolution inequality, with $r=p'$). $\quad\Box$