# If $\lim_\limits{n\to\infty}\int_0^1f(x+n)dx=2$, prove that $\lim_\limits{n\to\infty}\int_0^1f(nx)dx=2$.

It is given $$f:[0,\infty)\to\mathbb{R}$$ is a continuous function and $$\lim_{n\to \infty}\int_{0}^{1} f(x+n)dx=2$$ then prove that $$\lim_{n\to \infty}\int_{0}^{1} f(nx)dx=2$$.

I can prove that $$\lim_{x\to \infty} f(x)=2$$ using the mean value theorem.

But then i used M.V.T. in second integral and getting $$\lim_{n\to \infty}\int_{0}^{n} \frac{f(u)}{n}du=f(c)$$ for some $$c\in (0,n)$$, but i cant conclude that $$c\to\infty$$ and $$f(c)\to 2$$.

I have thought of using Lebesgue dominated convergence theorem and I have proven it but is there any other way to prove this with only elementary real analysis?

• I doubt that you can prove $\lim_{x \to \infty} f(x) = 2$. Consider for example $f(x) = 2 + \sin(2 \pi x)$. Commented Jun 15 at 13:05
• okay i didnt thought about this, using m.v.t. i think that was $c\in (n,n+1)\forall n\in\mathbb{N}$ such that $f(c)=2$. Commented Jun 15 at 13:33

## 1 Answer

Change of variables gives $$\int_0^1 f(x+n)dx=\int_n^{n+1}f(x)dx$$, and $$\int_0^1 f(nx)dx=\frac{1}{n}\int_0^n f(x)dx$$.

So if we define the sequence $$a_n=\int_{n-1}^n f(x)dx$$ then we have $$\int_0^1 f(nx)dx=\frac{a_1+...+a_n}{n}$$. And it is a very standard result about limits that if $$a_n\to L$$ then its sequence of arithmetic means tends to $$L$$ as well.

• Indeed, that is the Stolz–Cesàro theorem Commented Jun 15 at 12:39
• @MartinR The Stolz–Cesàro theorem is more general. I think that this particular case is due to Cesàro. Commented Jul 5 at 18:56