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](https://www.sciencedirect.com/science/article/abs/pii/S0022039623002255) 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!



Does $c$ in the embedding inequality $\|P^\kappa_t \|_{L^p \to L^{p'}} \le c t^{-\frac{d(p'-p)}{2pp'}}$ depend on $\kappa$?
$\newcommand\ka\kappa$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](https://en.wikipedia.org/wiki/Young%27s_convolution_inequality#Euclidean_space), with $r=p'$). $\quad\Box$


