Let $f(x)$ be integrable, periodic with period $l$, and integrate to $A$ over one period
Define $f_n(x) := f(nx)$
Show that $f_n(x)$ converges to $f(x)=\frac{A}{l}$ in the sense of distributions
I've been trying to adapt a similar proof for $\sin(nx)$, but $\sin$ has properties not possessed by all periodic functions so it doesn't work.
We want to show that $\forall\epsilon>0 \ \exists N\in\mathbb{N}$ such that if $n>N$ then
$$\left| \int_{-\infty}^{\infty}f_n(x)\phi(x) \ dx \ - \ \int_{-\infty}^{\infty}\frac{A}{l}\phi(x) \ dx\right| \ < \epsilon$$
My first instinct is to combine the integrals and bound it using tools such as the Mean Value Theorem and the Triangle Inequality, but everything I try hits a roadblock.
Any help is greatly appreciated!
Let $g(x)= f(x)-A/l$ and $G(x) = \int_0^x g(y)dy$, integrate by parts $$\int_{-\infty}^\infty g(nx) \phi(x)dx$$ and use that $\frac{G(nx)}{n} = O(\frac1n)$ and $\phi'\in L^1$