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I would like to show that given a function $f:\mathbb{R}\to \mathbb{R}$ such that is continuos, $|f(x)| < M$ and $\exists z \in \mathbb{R} $ such that $$-\infty < \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(z + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon < +\infty$$ then:

$$ \forall x,y \in \mathbb{R} \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(x + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(y + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon $$

I started from the following:

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

then we want to show that

$$\lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} (f(y+\varepsilon)-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon$$

Let $\delta$ be arbitrary then there is $\theta = \theta_{\delta} > 0$ such that

$$\int_{-\theta}^{+\theta} (f(y+\varepsilon)-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon < \int_{-\theta}^{+\theta} (M-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon$$

(\varepsilon is bounded by -\theta and +\theta) and

$$\lim_{\sigma \to +\infty} \int_{-\theta}^{+\theta} (M-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon = 0$$

Now It is more difficult to show that the integral between $+\theta$ and $+\infty$ (or the integral between $-\infty$ and $-\theta$ goes to 0. I think in someway the fact that with $x$ it was bounded.

Sam
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    Without further conditions on $f$ the limit need not exist. If one of them exists, however, then so does the other and they are equal. – H. H. Rugh Sep 26 '16 at 11:47
  • You mean for example that $f$ should be such that $\lim_{x \to +\infty} f(x) = c_1$ and $\lim_{x \to -\infty} f(x) = c_2$ – Sam Sep 26 '16 at 11:54
  • If $\int_{-\infty}^\infty f^2(x){\rm d}x < \infty$ then you can use Cauchy-Schwarz to show that the limit has to be zero. – Winther Sep 26 '16 at 11:55
  • @Rugh It is interesting that you say that if one exist then the other exist and they are the same. I think will be nice for me to show this. And I have no idea where to start. Then I could also show that with some more conditions on $f$, than the limit exist so just I show tha it could happen that the limit exists. – Sam Sep 26 '16 at 12:02
  • @Winther thanks it could be useful (even if I'm not sure I will be able to use Cauchy-Schwarz - I'm not a matematician, I just try) – Sam Sep 26 '16 at 12:05
  • Important to choose a good cut-off. As I indicated below, $\theta =\sigma^{3/2}$ should do. – H. H. Rugh Sep 26 '16 at 16:04

1 Answers1

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In the present context continuity of $f$ is not important. Rather what happens at $\pm \infty$. Setting $$\rho_\sigma(\varepsilon) = \frac{1}{\sqrt{2\pi} \sigma} \exp\left( - \frac{\varepsilon^2}{\sigma^2}\right) $$

We consider your difference for fixed $x\neq 0$ and with $y=0$: $$ \int_{\Bbb R} \left(f(\varepsilon+x) -f(\varepsilon)\right) \rho_\sigma(\varepsilon) \; d\varepsilon = \int_{\Bbb R} f(\varepsilon) \left( \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon ) \right)\; d\varepsilon$$ We want to show it tends to zero which implies the claim. We have $$ \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon) = \rho_\sigma(\varepsilon) \left[ \exp\left(\frac{2\varepsilon x -x^2}{2\sigma^2}\right)-1 \right] $$ Let $1>\delta>0$ and consider $\epsilon<\sigma^{3/2}$. Then the argument of the exponential is bounded by $$ \left|\frac{2\varepsilon x -x^2}{2\sigma^2}\right|\leq \frac{|x|}{\sqrt{\sigma}} + \frac{x^2}{2\sigma^2}=: a_\sigma$$ which goes to zero as $\sigma\rightarrow \infty$. This implies that $$\int_{-\sigma^{3/2}}^{\sigma^{3/2}}| f(\varepsilon) \left( \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon ) \right)|\; d\varepsilon \leq M (\exp(a_\sigma)-1) \rightarrow 0$$ as $\sigma\rightarrow \infty$. For the integral from e.g. $\sigma^{3/2}$ to infinity one shows in a similar way (leaving out the details..., $\rho_\sigma(\varepsilon)$ will be 'very small') that it goes to zero as $\sigma\rightarrow \infty$ whenever $f$ is bounded.

H. H. Rugh
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