# $\frac{\partial}{\partial x}\int_{\mathbb{R}}\frac{1}{\sqrt{2 \pi \varepsilon}}e^{-\frac{(x-y)^2}{2\varepsilon}}l(y)dy\leq C\frac{1}{x}$

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 $$\varepsilon$$.

• Doesn't hold. For $l(x)=\cos x$ the integral is equal to $e^{-\varepsilon/2} \cos (x)$. – Andrew May 10 at 7:53
• Maybe $C/\sqrt{\epsilon}$. – Giorgio Metafune May 10 at 8:01
• No I don't want $C/\sqrt{\epsilon}$. – yassine yassine May 10 at 8:07

## 1 Answer

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.