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Alessio Rondelli 2024-11-19 11:07:23 +01:00
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@ -178,7 +178,8 @@ Let's first tackle the case with $f$ Lipschitz. Let $x\in\R^N$. We have
The second inequality for the Lipschitz case is done in a completely analogous manner. Let's consider the Holder case
\end{comment}
\begin{align*}
\left| \partial_{x_ix_j}V^{s,t}_\nu f (x) \right|&=\left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)f(y)dy \right|\leq \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)(f(y)-f(e^{(t-s)B}x))dy \right|+ \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)dyf(e^{(t-s)B}x) \right|\\
\left| \partial_{x_ix_j}V^{s,t}_\nu f (x) \right|&=\left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)f(y)dy \right| \\
&\leq \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)(f(y)-f(e^{(t-s)B}x))dy \right|+ \left| \int\partial_{x_i x_j}p^\nu(s,x;t,y)dyf(e^{(t-s)B}x) \right|\\
&\leq \int |\partial_{x_ix_j}p^\nu(s,x;t,y)||e^{(t-s)B}x-y|^\alpha_B dy + \left| \partial_{x_ix_j}\underbrace{\int p^\nu(s,x;t,y) dy}_{=1} \right||f(e^{(t-s)B}x)|\\
&\leq \frac{C_{B,\alpha}}{|t-s|}\int \Gamma^+(t-s,x-y)|x-e^{-(t-s)B}y|^\alpha_B dy + 0\leq \frac{C_{B,\alpha}}{|t-s|^{1-\frac{\alpha}{2}}}.
\end{align*}
@ -219,7 +220,7 @@ Now if we consider $d_B(x,y)=|x-y|_B$ and $K^\epsilon=\left\lbrace x\in\R^N\ |\
$$
\sup_{x\in K^\epsilon}|f(x)-f_j(x)|\leq \sup_{x\in K^\epsilon}\left( |f(x)-f(y_x)| + |f(y_x)-f_j(y_x)| + |f_j(y_x)-f_j(x)|\right)\leq \sup_{x\in K^\epsilon}\left( 2\epsilon^\alpha + \epsilon \right)\leq C_\alpha\epsilon^\alpha.
$$
where $y_x$ is a point in $K$ such that $|x-y|_B<\epsilon$. $C_\alpha$ may be taken uniformly of $\epsilon$ as long as $\epsilon\leq 1$.
where $y_x$ is a point in $K$ such that $|x-y_x|_B<\epsilon$. $C_\alpha$ may be taken uniformly of $\epsilon$ as long as $\epsilon\leq 1$.
Let $g(x)=\max\left( 0, 1-\frac{d_B(x,K)}{\epsilon} \right)$, evidently $g\in bLip\subseteq C^{\alpha}_B$ and $\mathds{1}_K\leq g\leq \mathds{1}_{K^\epsilon}$. Thus by taking $n$ big enough we have by convergence against $C^{\alpha}_B$ functions that
$$