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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following problem:

Let $\phi(x) \in C_0^\infty(\mathbb{R}^n)$ satisfy $\phi \geq 0$ and $\phi(0) = 1$. Show that $\phi_n = \phi/n$ converge to $0$ in $\mathcal{D}'(\mathbb{R}^n)$.

My solution sketch is the following. $\lim_{n->\infty} \int_{\mathbb{R}^n} \frac{\phi(x)}{n} f(x)dx = \lim_{n->\infty} \frac{1}{n}\int_{\mathbb{R}^n} \phi(x) f(x)dx \leq \lim_{n->\infty} \frac{C}{n} = 0$ where I in the last step have used that $\phi$ has compact support so that the integral will be finite.

I'm a little confused about this solution since I have not used the addional properties of $\phi$ and therefore I think I have cheated a little. So my question is then what I have missed in my solution?

Thanks in advice!

share|cite|improve this question
You just proved that the limit is less or equatl to zero. You still have to finish the proof. – Tomás Feb 14 '13 at 15:53
I missed that, sorry. But anyhow, with the same reasoning as before I can also say that the integral will be bigger than another constant, since it finite, right? – Nils Feb 15 '13 at 8:02
Yes, I think so, and looks like the hypothesis $\phi(0)=0$ and $\phi\geq 0$ are useless. Mayve it is $\phi_n(x)=\phi(\frac{x}{n})$. What do ypu think? – Tomás Feb 15 '13 at 11:18
Ok, thanks. I suspected that the hypothesis was there because of that this exercise was once ago a bigger one, but it made me a little bit confused. – Nils Feb 15 '13 at 13:02

If $\phi_n(x)=\frac 1n\phi(x)$, then just notice that if $\varphi\in\mathcal D(\Bbb R)$, $$\int_{\Bbb R}\varphi_n(x)\varphi(x)dx=\frac 1n\int_{\Bbb R}\phi(x)\varphi(x)dx,$$ and $\int_{\Bbb R}\phi(x)\varphi(x)dx$ is a real number (since $\varphi$ and $\varphi$ have a bounded support and are bounded). So the limit as $n\to \infty$ is $0$. (the other assumptions are superfluous)

If $\phi_n(x)=\phi\left(\frac xn\right)$, then write $$\phi_n(x)=1+\int_0^x\frac 1n\phi'\left(\frac tn\right)dt,$$ which gives that $$\int_{\Bbb R}\phi_n(x)\varphi(x)dx=\int_{\Bbb R}\varphi(x)dx+\int_{\Bbb R}\varphi(x)\int_0^x\frac 1n\phi'\left(\frac tn\right)dtdx,$$ and the inequality $$\left|\int_0^x\frac 1n\varphi'\left(\frac tn\right)dtdx\right|\leqslant \frac{|x|}n\sup_{\Bbb R}|\phi'|\cdot\chi_{\operatorname{Supp}\phi},$$ hence $\phi_n\to 1$ in $\mathcal D'(\Bbb R)$.

share|cite|improve this answer

Your Answer


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.