An integrable estimate of the Hölder constant of the map $x \mapsto \int_{\mathbb R^d} f(y) \partial_1 \partial_1 g_t (x-y) \, \mathrm d y$

Question

Let $(g_t)_{t>0}$ be the Gaussian heat kernel on $\mathbb R^d$, i.e., $$ g_t (x) := (4\pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{4t}}, \quad t>0, x \in \mathbb R^d. $$

Let $f : \mathbb R^d \to \mathbb R$ belong to the Hölder space $C^{0, \alpha}_b (\mathbb R^d)$ for some $\alpha \in (0, 1)$. Let $[f]_\alpha$ be the best $\alpha$-Hölder constant of $f$, i.e., $$ [f]_\alpha := \sup_{x \neq y} \frac{|f(x) - f(y)|}{|x-y|^\alpha}. $$

We define $F_t : \mathbb R^d \to \mathbb R$ by $$ F_t(x) := \int_{\mathbb R^d} f(y) \partial_1 \partial_1 g_t (x-y) \, \mathrm d y, $$ where $\partial_1 := \frac{\partial}{\partial x_1}$ is the partial derivative.

Is there an integrable function $c:(0, 1) \to \mathbb R$ such that $[F_t]_\alpha \le c(t) [f]_\alpha$ for all $t \in (0,1)$?

Thank you so much for your elaboration!

Answer 1

If $T_tf=g_t*f$ is the heat semigroup you are asking for the norm of $D_{ij}T_t$ from $C^\alpha$ to itself (endowed with the Holder seminorm).

Let $I_\lambda f(x)=f(\lambda x), \lambda >0$. Then $[I_\lambda f]_\alpha=\lambda ^\alpha [f]_\alpha$ and $T_t=I_{\frac{1}{\sqrt t}} \circ T_1 \circ I_{\sqrt t}$, by explicit computation or the scaling properties of the heat equation.

This gives $D_{ij}T_t=\frac{1}{t}I_{\frac{1}{\sqrt t}} \circ D_{ij} T_1 \circ I_{\sqrt t}$ and the norm of $D_{ij}T_t$ is $t^{-1}$ times the norm of $D_{ij}T_1$.

No integrable $c(t)$ exists and this explain why Schauder estimates are not so easy to proof; same argument for $L^p$ estimates.