Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 at 0:28
    
@Seyhmus: Yes it is not compactly supported. –  Paulo Feb 28 at 0:49

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
Thanks for this wonderful proof, that is exactly what I was asking for ! –  Paulo Feb 28 at 0:41
1  
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 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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.