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Let $l$ be a continuous bounded function ($l$ is not differentiable). I want to prove for $x$ large enough that $$\frac{\partial}{\partial x}\int_{\mathbb{R}}\frac{1}{\sqrt{2 \pi \varepsilon}}e^{-\frac{(x-y)^2}{2\varepsilon}}l(y)dy\leq C\frac{1}{x}. $$ Where $C$ is a positive constant, and uniformly in $\varepsilon$.

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    $\begingroup$ Doesn't hold. For $l(x)=\cos x$ the integral is equal to $e^{-\varepsilon/2} \cos (x)$. $\endgroup$ – Andrew May 10 at 7:53
  • $\begingroup$ Maybe $C/\sqrt{\epsilon}$. $\endgroup$ – Giorgio Metafune May 10 at 8:01
  • $\begingroup$ No I don't want $C/\sqrt{\epsilon}$. $\endgroup$ – yassine yassine May 10 at 8:07
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This cannot hold true, at least not without further assumptions on $l$ (more precisely, on the decay of $l'(x)$ at infinity). Indeed, let $l_\epsilon:=\Gamma_\epsilon \ast l$ be the $\epsilon$-mollification as in your statement (here $\Gamma_\epsilon$ is the heat kernel at time $\epsilon$). You are asking whether you can contol $|l_\epsilon'(x)|\leq \frac{C}{x}$ for large $x$, uniformly in $\epsilon$. But it is well-known that, if for example $l$ is smooth enough (say $C^1$) then $l_{\epsilon}'=\Gamma_\epsilon\ast (l')$, which converges at least pointwise to $l'$ as $\epsilon\to 0$. So, roughly speaking, if $l'$ does not decay at infinity at least as $\frac{1}{|x|}$ you cannot expect this estimate to be true.

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