$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bF}{\mathbb{F}} \newcommand{\cF}{\mathcal{F}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\qtext}[1]{\quad\text{#1}} \newcommand{\qtextq}[1]{\quad\text{#1}\quad} \DeclareMathOperator*{\trace}{tr} \DeclareMathOperator*{\Div}{div} $ We fix $T \in (0, \infty)$ and let $\bT$ be the interval $[0, T]$. We consider the measurable function \begin{align} \sigma &: \bT \times \bR^d \to \bR^d \otimes \bR^m . \end{align}
Let $a := \sigma \sigma^\top$. We denote
- $\sigma_t := \sigma (t, \cdot)$ and $a_t := a (t, \cdot)$.
- by $\nabla$ the gradient operator w.r.t space variable.
- by $\partial_t$ the derivative w.r.t time.
We fix $\alpha \in (0, 1)$. We assume there exists a constant $C >0$ such that for $t \in \bT$ and $x, y \in \bR^d$:
- $a_t$ is invertible and $\sigma_t$ is differentiable.
- $\frac{|\nabla \sigma_t (x) - \nabla \sigma_t (y)|}{|x-y|^\alpha} + | \nabla \sigma_t (x) | + | \sigma_t (x) | + | a_t^{-1} (x) | \le C$.
We fix a probability density function $f : \bR^d \to \bR_+$ such that $$ \|f \|_{C^\alpha_b} := \| f \|_\infty + \sup_{x, y \in \bR^d; x \neq y} \frac{|f (x) - f (y)|}{|x-y|^\alpha} \le C. $$
Let $u : \mathbb T \times \mathbb R^d \to \mathbb R_+$ be the distributional solution to the Cauchy problem $$ \begin{cases} \partial_t u (t, x) &= \Div \{ a (t, x) \nabla u(t, x) \} , \\ u(0, \cdot) &= f . \end{cases} $$
I wonder if there exists a constant $c$ (depending only on $\alpha, C, T, d$) such that $$ \sup_{t \in \bT} \| u(t, \cdot) \|_{C^\alpha_b} \le c \|f \|_{C^\alpha_b} . $$
Thank you very much for your elaboration. Any reference is greatly appreciated.
