Suppose $f:[0,\infty)\rightarrow [0,\infty)$ is integrable. Set $f_{n}(x):=f(nx)$. I want to show that $f_{n}(x)\rightarrow 0$ almost everywhere or equivalently, the set $$\{x : \limsup_{n}f_{n}(x)\geq\delta\}$$ has measure zero, for any $\delta>0$.

By dilation invariance, it's clear that $f_{n}\rightarrow 0$ in $L^{1}$ and therefore also in measure. Furthermore, we can pass to a subsequence to obtain a.e. convergence. If $f$ has compact support, then it's obvious that $f_{n}\rightarrow 0$ almost everywhere.

My thought was to try approximating $f$ in $L^{1}$ by $g\in C_{c}(\mathbb{R})$ and use something like

$$|\{\limsup f_{n}\geq\delta\}|\leq|\{\limsup|f_{n}-g_{n}|\geq\delta/2\}|+|\{\limsup|g_{n}|\geq\delta/2\}|$$

and go from there. But I'm not sure how to control the first term on the RHS. Any suggestions?


Fix $a,b$ with $0<a<b$. It's enough to show that $$\int_a^b\sum_nf_n<\infty.$$ But $\int_a^bf_n=\frac1n\int_{na}^{nb}f$, so $$\int_a^b\sum_nf_n=\int_a^\infty\phi f,$$where $$\phi=\sum\frac1n\chi_{(na,nb)}.$$So it's enough to show that $\phi$ is bounded on $(a,\infty)$.

And $\phi$ is certainly bounded on $(a,R)$ for any $R<\infty$, so we need only get a bound on $\phi(x)$ for large $x$. Now, $$\phi(x)=\sum_{n\in\left(x/b,x/a\right)}\frac1n.$$At least for large $x$ it seems clear that this is bounded by something like $$O(1)+\int_{x/b}^{x/a}\frac{dt}{t}=O(1)+\log(b/a)=O(1).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.