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I have some troubles with the following problem:

A definition

Let $\sigma_1$ and $\sigma_2$ two positive numbers. We denote for all $x\in\mathbb{R}$, >$G_\sigma\left[ \phi \right](x)$ the gaussian convolution (with std $\sigma>0$) of a function $\phi$ at point $x$, in other words:

$$\forall x \in \mathbb{R},~ G_\sigma\left[ \phi \right](x) :=\frac{1}{\sqrt{2\pi}\sigma} \int_\mathbb{R}\phi(x-s)e^{-\frac{s^2}{2\sigma^2}}ds.$$

My question

Let $f$ and $g$ be two continuous functions on $\mathbb{R}$. Is there a way to bound the real number

$$\Delta_{\sigma_1,\sigma_2}[f;g]:= \|G_{\sigma_1}[f]-G_{\sigma_2}[g]\|_\infty$$

with a something depending on $\|f-g\|_\infty$.

Illustration of my post

What I would like is something like:

$$\exists (a,b,k)\in\mathbb{R}^3,~\forall x \in \mathbb{R}, \Delta_{\sigma_1,\sigma_2}\left[f;g\right](x) \leq a+b\|f-g\|_\infty^k$$

with constants that can depend on $\sigma_1$ and $\sigma_2$.

Any hint would be highly appreciated, thank you!

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    $\begingroup$ (i) Is it $e^{s^2/(2\sigma^2)}$ or $e^{-s^2/(2\sigma^2)}$? (ii) The "pointwise" inequality you want is quite impossible. Indeed, it is easy to get an example when (say) $f(0)=g(0)$, but the left-hand side of your desired inequality is $>a$, whatever $a$ was. $\endgroup$ Commented Jan 15 at 15:43
  • $\begingroup$ Thank you @IosifPinelis, I have edited the post. I didn't mention that the constants are allowed to depend on $\sigma_1$ and $\sigma_2$, do you think what you said still holds in this case ? (It was badly formulated...) $\endgroup$
    – NancyBoy
    Commented Jan 15 at 15:47
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    $\begingroup$ What I said holds for any given $\sigma_1$ and $\sigma_2$. So, allowing the "constants" to depend on $\sigma_1$ and $\sigma_2$ does not change anything. $\endgroup$ Commented Jan 15 at 16:02
  • $\begingroup$ Ok, I see, I have edited it with infinite norm. It is maybe doable with this. Thank you again @IosifPinelis. $\endgroup$
    – NancyBoy
    Commented Jan 15 at 16:05
  • $\begingroup$ Now your inequality is trivial, with $a=0$, $b=1$, $k=1$. $\endgroup$ Commented Jan 15 at 16:06

1 Answer 1

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$\newcommand\si\sigma$Such a bound does not exist. E.g., suppose that $f(u)=g(u)=cu^2$ for some real $c$ and all real $u$. Then your inequality becomes $$c|\si_1^2-\si_2^2|\le a,$$ which will not hold if $\si_1\ne\si_2$ and $c>a/|\si_1^2-\si_2^2|$.

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  • $\begingroup$ Thank you for this great counter-example ! Just for intuition, what dependence on the constants could make this inequality hold ? $\endgroup$
    – NancyBoy
    Commented Jan 15 at 16:39
  • $\begingroup$ @NancyBoy : A valid bound should also depend on $\|f\|_\infty$, $\|g\|_\infty$, and the moduli of continuity of $f$ and $g$. $\endgroup$ Commented Jan 15 at 17:26

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