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

Question

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?

Answer 1

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.