# Gaussian function as a mollifier?

I am wondering if the function $f_{n}(x)=\sqrt{\frac{1}{2\pi n^2}} exp(\frac{-x^2}{2n^2})$ can be viewed as a mollifier. I would like to prove that if $g$ is a continuous and bounded function then $f_{n}*g$ converges to $f*g$ as $n\rightarrow 0$ ( if $f_n$ is mollifier then the result is straighforward). I know that $f_n$ is a density so its integral over $\mathbb R$ is equal to 1, I guess that this function is smooth ($\mathcal{C}^{\infty}$). An clearly $f_n$ goes to zero when $\vert x \vert \rightarrow +\infty$, but I don't have the hypothesis that $f_{n}$ is compactly supported... thus is not clear for me whether $f_n$ is a mollifier... any suggestion?

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The support of $f_n(x)$ is $\mathbb{R}$, which is not compact, therefore $f_n$ is in a wide sense not a mollifier. – Seyhmus Güngören Feb 28 '14 at 0:28
@Seyhmus: Yes it is not compactly supported. – Paulo Feb 28 '14 at 0:49

Good question. Note $f_n(x)=1/n f(x/n)$ where $f(x)=\frac{1}{2 \pi} e^{-x^2/2}$. Approximate $f(x)$ by $h^\epsilon(x)$ in the $L^1$ norm such that $h^\epsilon(x)$ is positive and compactly supported (for instance, take $h^\epsilon(x)=\phi(x/\epsilon) f(x)$ where $\phi$ is a positive smooth bump function supported in $[-1,1]$).

Now do all the necessary estimates. That is, write

$$|(f_n * g)(x)-g(x)|\leq |((f_n-h^\epsilon_n)*g_n)(x)|+|(h^\epsilon_n * g_n)(x)-g(x)|$$ and then by Young's inequality with another triangle inequality $$|(f_n * g)(x)-g(x)|\leq \epsilon ||g||_\infty+|(h^\epsilon_n * g_n)(x)-||h^\epsilon||_{L^1} g(x)| + |||h^\epsilon||_{L^1}-1|\cdot|g(x)|.$$

Here's the key step. Since $h^\epsilon$ is compactly supported, dividing out by it's $L^1$ norm makes it a mollifier, and so $\frac{1}{||h^\epsilon_{L^1}||} h^\epsilon_n * g \to g$ as $n \to \infty$. So take the limsup of the above with respect to $n$ to get

$$\limsup_{n \to \infty} |(f_n * g)(x)-g(x)|\leq \epsilon ||g||_\infty+ |||h^\epsilon||_{L^1}-1|\cdot|g(x)|.$$

Now take $\epsilon \to 0$, and note that $||h^\epsilon|| \to 1$ to get that $$\limsup_{n \to \infty} |(f_n * g)(x)-g(x)| \leq 0,$$ hence the limit exists, hence we have convergence.

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Thanks for this wonderful proof, that is exactly what I was asking for ! – Paulo Feb 28 '14 at 0:41
Remember in analysis: always think to approximate, always think to estimate (bound above), and always think to isolate (split up between the good and bad sets). – nayrb Feb 28 '14 at 1:18

Smoothing a function by convolution with $f_n$ is a classical technique. Compared to mollifiers (with compact support) a big advantage is that $f_n$ and hence $f_n * g$ are not only smooth but analytic. Cutting off the Taylor series of the convolution you then get even polynomial approximation on compact sets, that is, the theorem of Weierstraß.

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Could you go a bit more into detail here? Why is it "classical", means, in which book can I find it? – tomglabst Oct 23 '14 at 15:29