Let $f\in L^1(R)$ set $f_n(x)=\dfrac{f(nx)}{n}$ prove that $\lim_{n \to \infty}f_n=0$ for a.e x

Question

Let $f\in L^1(R)$, set $f_n(x)=\dfrac{f(nx)}{n}$, $n\geq 1$ prove that $\lim_{n \to \infty}f_n=0$ for almost every $x\in R$

what I am thinking.

Since $f\in L^1(R)$ ie. $\int_R |f|<\infty$ it implies $f$ is finite a.e on $R$.

So by possibly excising a set $E$ of measure zero, $f $ becomes finite everywhere. hence $\exists M \geq 0$ such $|f(x)|\leq M, \forall x \in R-E$

therefore $|f_n(x)| \leq \dfrac{M}{n}$ for all n, implying $ \lim_{n \to \infty}f_n(x)=0$

It seems pretty straight forward unless I thinking about it wrongly. can someone please take a look ant let me know. thank you.

Answer 1

You have to be a little careful because the fact that $f$ is finite almost everywhere does not imply that $f$ is bounded almost everywhere.

Take for instance $f(x)=\frac{1}{x}$ on $(0,1]$

Now, do the substitution $u=nx$ and you will have:

$$\int f_n(x)dx=\frac{1}{n^2}\int f(u)du$$

Now we have that $$\sum_{n=1}^{\infty}\frac{1}{n}\int|f(nx)|dx=\sum_{n=1}^{\infty}\frac{1}{n^2}\int|f(u)| < \infty$$ because $f$ is Lebesgue integrable.

By the monotone convergence theorem we have that $\int \sum_{n=1}^{\infty}\frac{1}{n}f(nx)dx< \infty$

Thus $\sum_{n=1}^{\infty}\frac{1}{n}f(nx) < \infty $ almost everywhere therefore $\frac{1}{n}f(nx) \to 0$ almost everywhere.