Show that $\lim_{\sigma \to 0} \int{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\epsilon} = f(x)$

Question

I would like to show that

$$\lim_{\sigma \to 0} \int{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon} = f(x)$$

This is reasonable because $\int{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}}$ is the average of $f(x+\varepsilon)$ weighted with $\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}$, and closing $\sigma$ to be $0$ we give more importance to $f(x+\varepsilon)$ when $\varepsilon$ is close to $0$. I apologize if my explanation lack of formality but I just studied Software Engineering, not math. At the moment I consider $\varepsilon, \sigma, x$ to be pure numbers, not vectors or matrixes.

By the way, I started to think to divide the integral in three different component: the first one from $-\infty$ to $-0$, the second one from $-0$ to $+0$, and the third one from $+0$ to infinite, and I would like to show that if $\sigma \to 0$ the second integral will become $f(x)$ and the first and the third one will become 0.

In other words

$$\lim_{\sigma \to 0}\lim_{\theta \to 0}\left[ \int_{-\infty}^{-\theta}{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon}+ \int_{-\theta}^{+\theta}{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^"}}\mathrm{d}\varepsilon}+ \int_{+\theta}^{+\infty}{f(x+\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon}\right] $$

I started from the integral in the middle. I thought that I could write

$$\lim_{a \to 0} \int_{-a}^{+a}{g(x)\mathrm{d}x} = \int_{-a}^{+a}{ \lim_{x \to 0} g(x) \mathrm{d}x}$$

And actually I don't know if this is correct or not. By the way I did the following steps for the integral in the middle:

$$\lim_{\theta \to 0} \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\theta}^{+\theta} f(x+\varepsilon) \left({e^{-\varepsilon^2}}\right)^{\frac{1}{2\sigma^2}}\mathrm{d}\varepsilon$$

$$\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\theta}^{+\theta}\lim_{\varepsilon \to 0} f(x+\varepsilon) \left({e^{-\varepsilon^2}}\right)^{\frac{1}{2\sigma^2}}\mathrm{d}\varepsilon$$

$$\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\theta}^{+\theta}\lim_{\varepsilon \to 0} f(x+\varepsilon) \left(-\varepsilon^2+1\right)^{\frac{1}{2\sigma^2}}\mathrm{d}\varepsilon$$

$$\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\theta}^{+\theta} f(x+\varepsilon) \left(-\frac{\varepsilon^2}{2\sigma^2}+1\right)\mathrm{d}\varepsilon$$

the last step is provided because $\lim_{x \to 0} \frac{(1+x)^{\alpha}}{x} = \alpha$ and so $\lim_{x \to 0} (1+x)^{\alpha} = \alpha x + 1$ I continued as the following:

$$\frac{1}{2\sqrt{2\pi}\sigma^3} \int_{-\theta}^{+\theta} -f(x+\varepsilon) \varepsilon^2\mathrm{d} \varepsilon + \frac{1}{\sqrt{2\pi}\sigma} F(x+\varepsilon)$$

($\sigma > 0 $) Now I don't know how to continue. I would like to show that:

$$\lim_{\sigma \to +0} \frac{1}{2\sqrt{2\pi}\sigma^3} \int_{-\theta}^{+\theta} -f(x+\varepsilon) \varepsilon^2\mathrm{d} \varepsilon + \frac{1}{\sqrt{2\pi}\sigma} F(x+\varepsilon) = f(x)$$

And then I think I know how to show that the integrals from $-\infty$ to $-\theta$ and the one from $+\theta$ to $+\infty$ are going to 0.

I'm sorry if my explanation it is very bad or if I made some terrible mistake. I'm not a matematician, but I would like to learn ho to become :). P.s if you can suggest me some tag.

Answer 1

Hints: It is convenient (if not necessary) to assume $f$ continuous. Also better assume that $f$ is bounded (although less may do). Your constants are not placed correctly. They should be such that when $f$ is constant you get that constant. Your idea of splitting up the integral is fine. You may do that on the difference between the two sides:

$$ \frac{1}{\sqrt{2\pi} \sigma}\int \left(f(x+\epsilon) -f(x)\right) \ e^{-\frac{\epsilon^2}{2\sigma^2}} d\epsilon $$

The integral from $-\theta$ to $\theta$ becomes small when $\sigma\rightarrow 0$ because of the continuity of $f$. And the other terms because the exponential becomes very small.

Answer 2

Thanks to @H.H.Rough now I can answer my own question, for those ones that are slow like me and needs more passages.

[Some edits by H.H.Rugh in the following:]

We assume $f:{\Bbb R} \rightarrow {\Bbb R}$ continuous. We also assume that $\forall a \in \mathbb{R} :|f(a)| \leq M <+\infty$.

We want to show that $$\lim_{\sigma \to +0} \int_{-\infty}^{+\infty} f(x+\varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{-\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = f(x)$$

For $g\in \mathbb{R}$ a constant we have $$\lim_{\sigma \to +0} \int_{-\infty}^{+\infty} g \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{-\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = g$$ because the integral over the density function makes 1. Thus we want to show that $$\lim_{\sigma \to +0} \int_{-\infty}^{+\infty} (f(x+\varepsilon) - f(x)) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{-\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = 0$$

Let $\delta>0$ be arbitrary. Then by continuity of $f$ there is $\theta=\theta_\delta > 0$ so that $|f(x+\varepsilon)-f(x)| < \frac{\delta}{2}$ for every $|\varepsilon|\leq \theta$. Because the normal density function is positive and the integral over its domain is one we have: $$0 \leq \int_{-\theta}^{+\theta} \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon \leq 1$$ and we deduce that: $$ \int_{-\theta}^{+\theta} \left| f(x+\varepsilon)-f(x)\right| \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \frac{\delta}{2}$$

Given the $\theta$ from above we may find $\sigma_\delta > 0$ small enough so that for every $0<\sigma<\sigma_\delta$ we have: $$ \int_{+\theta}^{+\infty} \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \frac{\delta}{8M}$$ Since $|f(x+\varepsilon)-f(x)| \leq 2M$ we obtain: $$\int_{+\theta}^{+\infty} \left| f(x+\varepsilon)-f(x)\right| \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon \leq 2M \int_{+\theta}^{+\infty} \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \frac{\delta}{4}$$

By symmetry we also have for such values of $\sigma$. $$\int_{-\infty}^{-\theta} \left|f(x+\varepsilon)-f(x)\right| \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathrm{d}\varepsilon < \frac{\delta}{4}$$

Combining the above we see that for every $0<\sigma<\sigma_\delta$ we have: $$\int_{-\infty}^{+\infty} \left|f(x+\varepsilon) - f(x)\right| \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{-\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon < \delta$$

As $\delta>0$ was arbitrary we conclude:

$$\lim_{\sigma \to +0} \int_{-\infty}^{+\infty} (f(x+\varepsilon) - f(x)) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{-\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = 0$$