# Distributional limit of $\phi/n$

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?

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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 $\mathcal D'(\Bbb R)$.