$f\in L^1\implies \lim_{n\rightarrow\infty}f(n^2 x)=0$ a.e. $x\in\mathbb{R}$?

Question

Can someone provide a hint for the following:

$f\in L^1(\mathbb{R})\implies \lim_{n\rightarrow\infty}f(n^2 x)=0$ a.e. $x\in\mathbb{R}$

I can't really make heads or tails of it. Something makes me think convergence in measure; or even a duality argument, but the quantification on $x$ for which this is satisfied is throwing me. I am not sure how it arises.

This is from an old qualifying exam.

Answer 1

Fix an $m\in\mathbb{Z}$, we have \begin{align*} \sum_{n\in\mathbb{N}}\int_{m}^{m+1}|f(n^{2}x)|dx&=\sum_{n\in\mathbb{N}}\int_{n^{2}m}^{n^{2}(m+1)}\dfrac{1}{n^{2}}|f(y)|dy\\ &\leq\sum_{n\in\mathbb{N}}\dfrac{1}{n^{2}}\cdot\int_{\mathbb{R}}|f(y)|dy\\ &<\infty, \end{align*} so \begin{align*} \int_{m}^{m+1}\sum_{n\in\mathbb{N}}|f(n^{2}x)|dx<\infty, \end{align*} this implies \begin{align*} \sum_{n\in\mathbb{N}}|f(n^{2}x)|<\infty \end{align*} for a.e. $x\in[m,m+1]$, which entails $\lim_{n\rightarrow\infty}f(n^{2}x)=0$.

Now varying $m\in\mathbb{Z}$ to obtain the result.

Answer 2

Here's an outline of the proof. First begin by showing that for $f \in L^1(\mathbb{R})$ it holds that $\int_\mathbb{R} f(tx)\;dm(x) = \frac{1}{t}\int_\mathbb{R}f(x)\;dm(x)$. There is a typical strategy to prove such results which I am going to present briefly:

1. Show it when $f$ is the characteristic function of some measurable set.
2. Using 1. show it for simple functions.
3. Show it for positive functions by approximating $f$ with simple functions.
4. Use the fact that $f = f^+ - f^-$ and the linearity of the integral to prove the general case.

Having proven that use Beppo-Levi theorem to see that: $$\int_\mathbb{R} \sum_{n=1}^{\infty} |f(n^2x)|\;dm(x) = \sum_{n=1}^{\infty}\int_\mathbb{R}|f(n^2x)|\;dm(x) = \sum_{n=1}^{\infty}\frac{1}{n^2}\cdot||f||_1 < \infty. $$

This implies that $\sum_{n=1}^{\infty}|f(n^2x)|$ converges almost everywhere and this concludes the proof.