Let $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $\displaystyle T > 0.$ For any real numbers $\displaystyle a<b$ evaluate $\displaystyle \lim_{n\to\infty} \int_a^b f(nx) \ dx.$
Let $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $\displaystyle T > 0.$ For any real numbers $\displaystyle a<b$ evaluate $\displaystyle \lim_{n\to\infty} \int_a^b f(nx) \ dx.$
Hi NonCommAlg.
A continuous function is integrable, and the integral over a bounded interval of an integrable periodic function is bounded. Hence, using the substitution $\displaystyle u=nx,$
$\displaystyle \lim_{n\,\to\,\infty}\int_a^bf(nx)\,dx$
$\displaystyle =\ \ \lim_{n\,\to\,\infty}\frac1n\int_{na}^{nb}f(u)\,du$
$\displaystyle =\ \ 0$ since the integral is bounded
Something tells me I may have done something wrong, because, well, surely it can’t be that simple …
I'd agree with that approach up to that point, but I wouldn't go on to say that the limit is 0, because the length of the interval (in the u-integral) is getting unboundedly long.
Let $\displaystyle f_{\text{av}} = \frac1T\int_0^Tf(x)\,dx$ be the mean value of f over one period. Then the interval [na,nb] consists of $\displaystyle \frac{n(b-a)}T$ subintervals of length T (not counting odd bits at the ends, which we can dispose of with epsilons). So $\displaystyle \frac1n\int_{na}^{nb}f(u)\,du\approx (b-a)f_{\text{av}}$. That's my candidate for the limit.
Opalg's candidate, which is $\displaystyle \frac{b-a}{T} \int_0^T f(x) \ dx,$ is the correct answer. here's a more detailed solution:
$\displaystyle \int_a^b f(nx) \ dx = \frac{1}{n} \int_{na}^{nb} f(u) \ du= \frac{1}{n} \left[\sum_{k=0}^{m-1} \int_{na+kT}^{na+(k+1)T} f(u) \ du + \int_{na+mT}^{nb} f(u) \ du \right],$ where $\displaystyle na+mT < b \leq na+(m+1)T. \ \ \ \ \ (1)$
but $\displaystyle \int_{na+kT}^{na+(k+1)T} f(u) \ du =\int_0^T f(u) \ du,$ since $\displaystyle T$ is the period of $\displaystyle f.$ thus: $\displaystyle \int_a^b f(nx) \ dx = \frac{m}{n} \int_0^T f(u) \ du + \frac{1}{n}\int_{na+mT}^{nb} f(u) \ du. \ \ \ (2)$
we also from (1) have that $\displaystyle \lim_{n\to\infty}\frac{m}{n}=\frac{b-a}{T},$ which completes the proof because $\displaystyle f$ is bounded, say by $\displaystyle K,$ and thus:
$\displaystyle \left| \frac{1}{n} \int_{na+mT}^{nb} f(u) \ du \right| \leq(nb-na-mT)\frac{K}{n}=\left(\frac{b-a}{T}-\frac{m}{n} \right) \frac{K}{T}.$ hence: $\displaystyle \lim_{n\to\infty} \frac{1}{n} \int_{na+mT}^{nb} f(u) \ du =0,$ and the result follows from (2).