Does Gaussian convolution respects order? 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 $g$.
 A: 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 $|x|<\delta$.
Then:
$$\begin{align}(h*p_\sigma)(0)&=\int_{\mathbb R} h(t)p_\sigma(-t)\,dt\\
&=\int_{|t|<\delta}  h(t)p_\sigma(-t)\,dt + \int_{|t|>\delta} h(t)p_\sigma(-t)\,dt.\end{align}$$
Now:
$$\left|\int_{|t|>\delta} h(t)p_\sigma(-t)\,dt\right|<p_\sigma(\delta)\|h\|_1$$
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:
$$H(\sigma)=(h*p_\sigma)(0)$$
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.
A: 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.
A: 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.
