Assume that we have two continuous integrable functions $f,g \in L^1(\mathbb{R})$ such that, for some $x_0 \in \mathbb{R}$, we have, $$f(x_0) \leq g(x_0) \; \; \; \; (1).$$

Now let us define the Gaussian function with variance $\sigma>0$, $$p_\sigma:x \mapsto \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{x^2}{2\sigma^2}}.$$

I would like to know to know if $(1)$ can imply \begin{equation} f\star p_\sigma (x_0) \leq g\star p_\sigma (x_0) \; \; \; \;(2).\end{equation}

My insight is that, when $\sigma$ vanishes, $ p_\sigma $ converges to a Dirac distribution and $f\star p_\sigma (x_0)$ converges to $f(x_0)$, and therefore (2) is true.

However, when $\sigma$ is large, $f\star p_\sigma$ will roughly be constant, and the local property $f(x_0) \leq g(x_0)$ is likely to be lost in the convolution.

Therefore, it may exist some kind of phase transition where, if $\sigma$ is smaller than a certain quantity $C_{f,g}$, then $(2)$ is true. I believe that, if this is the case, $C_{f,g}$ would be likely to depend on the regularity of $f$ and $g$.

  • $\begingroup$ It's only true if $f(x_0)<g(x_0)$. It is possible for $f(x_0)=g(x_0)$ but not have your equality - they are equal in the limit. $\endgroup$ – Thomas Andrews Jul 23 '16 at 13:29
  • $\begingroup$ You probably don't need the non-negative part. $\endgroup$ – Thomas Andrews Jul 23 '16 at 13:32
  • $\begingroup$ @Thomas I agree about the non-negativity being unnecessary, I will edit the question. However, I don't understand what you mean by "it is possible for $f(x_0)=g(x_0)$ but not have your equality - they are equal in the limit". $\endgroup$ – PAM Jul 23 '16 at 13:36
  • $\begingroup$ Let $g(x)=0$ for all $x$ and $f(x)=|x|$ for all $x$. Then $f(0)\leq g(0)$, but $(f*p_\sigma)(0)>0=(g*p_\sigma)(0)$ for all $\sigma>0$. $\endgroup$ – Thomas Andrews Jul 23 '16 at 14:25
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    $\begingroup$ It would help if you asked your question more clearly. You start by asking a question you know is false - "I would like to know if..." Without conditions on $\sigma$, you already argue that the answer is "No." You also seem to know that it is true for some interval of $\sigma$ values near $0$. So what are you really asking? $\endgroup$ – Thomas Andrews Jul 23 '16 at 14:36

You need $f(x_0)<g(x_0)$. If $f(x_0)=g(x_0)$ this is not true.

It is also not true, as you note, for all $\delta$.

Assuming $f(x_0)<g(x_0)$, We won't need $f,g$ positive, so we can reduce it to letting $h(x)=g(x)-f(x)$ and ask

If $h(x)$ is continuous and $L_1$ and $h(x_0)>0$, is it true for some $\sigma>0$ that $(h*p_\sigma)(x_0)>0$?

We can assume $x_0=0$.

Since $h$ is continuous, we can find $\delta>0$ so that $h(x)>h(0)/2$ for $|x|<\delta$.


$$\begin{align}(h*p_\sigma)(0)&=\int_{\mathbb R} h(t)p_\sigma(-t)\,dt\\ &=\int_{|t|<\delta} h(t)p_\sigma(-t)\,dt + \int_{|t|>\delta} h(t)p_\sigma(-t)\,dt.\end{align}$$


$$\left|\int_{|t|>\delta} h(t)p_\sigma(-t)\,dt\right|<p_\sigma(\delta)\|h\|_1$$

Also, is $0<\sigma<\delta$: $$\begin{align}\int_{|t|<\delta} h(t)p_\sigma(-t)\,dt&\geq \int_{|t|<\sigma} h(t)p_\sigma(-t)\,dt\\ &\geq \int_{|t|<\sigma}\frac{h(0)}{2}p_\sigma(\sigma)\,dt \\ &\geq 2\sigma \phi_\sigma(\sigma)\frac{h(0)}{2}\\ &=\frac{e^{-1/2}h(0)}{\sqrt{2\pi}} \end{align}$$

So, it is true if we pick $\sigma$ so that $0<\sigma<\delta$ and $p_\sigma(\delta)<\frac{e^{-1/2}h(0)}{\sqrt{2\pi}\|h\|_1}$. Then $(h*p_\sigma)(0)>0$.

There can be lots of transitions, but it is true that if you define:


The $H$ is continuous on $(0,\infty)$ and, if $h\in L_1$ is continuous, then $H$ is continuous, and $$\lim_{\sigma\to 0^+} H(\sigma)=h(0)\\ \lim_{\sigma\to\infty} H(\sigma)=\int_{\mathbb R} h(t)\,dt$$

This means that if $h(0)>0$ then $H(\sigma)>0$ on some interval $(0,\sigma_0)$.

This also means that if $\int_{\mathbb R}h(t)\,dt >0$ then $H(\sigma)>0$ on some interval $(\sigma_\infty,\infty)$.

But if $\int_{\mathbb R}h(t)\,dt =0$, then it is possible for $H(\sigma)$ to oscillate towards zero.


Your guess in not correct. WLOG assume $x_0 = 0$. Now consider two functions $f, g$ as in your assumptions such that $f(0) = g(0)$, but $f \ge g$ everywhere and $f(x) = 1+g(x)$ on $[1,2]$. Then for all $\sigma$ $$ f\ast p_\sigma(0) - g \ast p_\sigma(0)= (f-g) \ast p_\sigma(0) \ge \int_1^2 p_\sigma(y) dy > 0 \, . $$ If however $f(0) > g<0)$, then $f \ast p_\sigma(0) < g \ast p_\sigma(0)$ will indeed hold for small $\sigma$, as was already pointed out in one of the comments.

  • $\begingroup$ Thanks for this clarifying example. Do you think that it is possible to quantify the "small $\sigma$" condition ? Perhaps using the modulus of continuity of $f-g$ ? $\endgroup$ – PAM Jul 23 '16 at 13:56
  • $\begingroup$ It also depends on the magnitude of $\int_\mathbb{R} (f(x)-g(x))_+ dx$. $\endgroup$ – Hans Engler Jul 23 '16 at 14:05

Let $h(x)=-x^2+\varepsilon,\;\varepsilon>0$. Clearly, $h(0)>0$. The convolution at $0$ is $-\sigma^2+\varepsilon$. Since $\varepsilon$ is arbitrary this shows that however small $\sigma^2$ one can find a function with negative convolution.


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