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I'm reading Controlled Diffusion Process by N.V. Krylov. On page 87-88, in the proof of theorem II.6.1, it says the following: Let $\sigma(t,x)$ be a matrix of dimension $d\times d$, and let $b(t,x)$ be a $d$-dimensional vector. Assume that $\sigma(t,x)$, $b(t,x)$ are given for $t\geq 0$, $x \in E_d$ ($d$-dimensional Euclidean space), also, in addition, are bounded and Borel-measurable with respect to $(t,x)$. Also, let the matrix $\sigma(t,x)$ be positive definite, and moreover let $(\sigma(t,x)\lambda,\lambda)\geq\delta |\lambda|^2$ for some constant $\delta >0$ for all $(t,x), \lambda \in E^d$. Let $\sigma^n(t,x)= \sigma_{\epsilon_n}(t,x)$, $b^n(t,x)=b_{\epsilon_n}(t,x)$ be the mollifications of $\sigma$, $b$, where $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$, $\epsilon \neq 0$. Clearly $\sigma_n \rightarrow \sigma$, $b_n \rightarrow b$ almost surely as $n\rightarrow \infty$.

Then it says on the 6th line of page 88: Note that the derivatives of $\sigma_n$, $b_n$ are bounded for each $n$. Hence the functions $\sigma_n$, $b_n$ satisfy the Lipschitz condition.

I don't understand why the mollified functions have bounded derivatives. Please help.

Thanks!

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    $\begingroup$ If "mollified" means convolution with a smooth kernel with compact support, $\sigma_\epsilon:=\sigma*\phi_\epsilon$, then $\partial^\alpha \sigma_\epsilon =\sigma* \partial^\alpha \phi_\epsilon$ has bounded derivatives (though, of course, not uniformly wrto $\epsilon$). $\endgroup$
    – Pietro Majer
    Commented Nov 29, 2015 at 8:38
  • $\begingroup$ I already figured it out, but still thanks a lot! $\endgroup$
    – ghjdnkmttrasda
    Commented Nov 29, 2015 at 8:50

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