Show $\lim_{n\to\infty} n^p f(nx) = 0$ exists in the distributional sense Let $f\in C^\infty(\mathbb R)$ be periodic, with period $2\pi$ and have mean zero ($\int^{2\pi}_0 f(x)dx =0$).  Show that for any positive integer $p$ the following limit is valid in the distributional sense.
$$ \lim_{n\to\infty} n^p f(nx)= 0$$
What I have so far is that 
 $$   \int^\infty_{-\infty} n^p f(nx)\phi(x)dx=\sum^{\infty}_{k=-\infty}n^p\int^{2\pi k+2\pi}_{2\pi k} f(nx)\phi(x)dx  $$
Using $y=x-2\pi k$ and $dy=dx$, 
$$ =\sum^{\infty}_{k=-\infty}n^{p}\int^{2\pi}_{0} f(ny+2\pi k n)\phi(y+2\pi k )dy $$
by periodicity of $f$, this becomes
$$ =\sum^{\infty}_{k=-\infty}n^{p}\int^{2\pi n}_{0} f(ny)\phi(y+2\pi k )dy $$
Since $\phi\in C^\infty_0$, there exist integers $a,b$ such that $\phi(y+2\pi k )=0\forall k<a,k>b$
$$ =\sum^{b}_{k=a}n^{p}\int^{2\pi n}_{0} f(ny)\phi(y+2\pi k )dy $$
I am unsure how to proceed further.  I suspect a Fourier series is involved. 
 A: Have you tried to perform a change of variables $y=nx$?
You will obtain a series which is a function of $\phi$.  Then it is a matter of using that function $\phi\in C^{\infty}_c({\mathbb R})$.  (Though it might not be the best approach).
We could use a Fourier series expansion for $f$, right?  In this case, $c_0=0$ where $c_k=\frac{1}{\pi}\int_{-\pi}^{\pi} f(t) dt.$  It might work, and it holds for all $t\in {\mathbb R}$.
Let's not forget that, in this case, for any $p>0$ we have
$$\sum |k|^p |c_k|^2 < \infty.$$
I think that performing integration by parts $p+1$ times you manage to obtain a function $F(x)$ which satisfies the same as $f$, and the solution follows. Thus the factor $n^p$ is no longer trouble (using that $\phi$ is of compact support, of course).
We simply, for $p$ fixed, reduce to the case where $p=0$, which we have observed presents no trouble.

We may assume wlog that $\phi(x)$ has its support contained within an open interval within $[-\pi, \pi]$ (simply, choose $N\in {\mathbb N}$ big enough and take $f$ to be $2\pi N$-periodic.)
Now, write $$f(x)=\sum_{0\neq k\in {\mathbb Z}} c_k e^{ikt}.$$  The desired integral is of the form
$$\int_{-\pi}^{\pi} \sum c_k e^{ikt} \phi(x)\ dx.$$
Now, let us concentrate on the individual integral
$$\int_{-\pi}^{\pi} e^{iknt} \phi(x) \ dx,$$
and perform $p+1$ times integration by parts.  Use that $\phi^{(m)}(\pi)=\phi^{(m)}(-\pi)=0$ for all $m\geq 0$ (by construction), and there is no division by zero.  You obtain a denominator of the form $(nk)^{p+1}$, and the desired result follows.
The sum is perfectly OK, by the above.
To prove convergence in the space of tempered distributions, you integrate by parts between $2\pi m$ and $2\pi m$ for sufficiently big m, and will be able to use the topology on $S({\mathbb R})$ with the terms appearing (up to $p$ derivatives of $\phi$).
