I would like to proove somehow that if $$f(x) = \int_{-\infty}^{+\infty} g(x+\varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon$$ with $g$ continuos and limited, than $f$ will be its smoothed version. I need first of all a system to measure the smoothness of a function, and actually I didn't find any. So I thought that I have to define one: given $p \in \mathbb{R}$ I say that $f$ is smoother than $g$ in $p$ given $\sigma$ if given: $$\mu_{f,p,\sigma} = \int_{-\infty}^{+\infty} f(p+\varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon$$ $$\mu_{g,p,\sigma} = \int_{-\infty}^{+\infty} g(p+\varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon$$ it holds:$$\int_{-\infty}^{+\infty} \left(f(p+\varepsilon)-\mu_{f,p,\sigma}\right)^2 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \int_{-\infty}^{+\infty} \left(g(p+\varepsilon)-\mu_{g,p,\sigma}\right)^2 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon $$
and, more in general $f$ is smoother than $g$ if it holds:
$$\int_{-\infty}^{+\infty} \left(f(x+\varepsilon)-\mu_{f,x,\sigma}\right)^2 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \int_{-\infty}^{+\infty} \left(g(x+\varepsilon)-\mu_{g,x,\sigma}\right)^2 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon $$ for any $\sigma>0$ and for any $x \in \mathbb{R}$
Now the question: do you think that the statement above is true? Do you think that this way to measure the smoothness of a function is right, or do you know a better way to measure it (such that it will be possible and easy to show the first statement in this question)?