0
$\begingroup$

Let $f\in \mathcal R[0,1]$ , and $g:\mathbb R \to \mathbb R $ is continuous and periodic with period $1$ . Then is it true that

$\lim_{n \to \infty}\int_0 ^1 f(x)g(nx)dx=\Big(\int_0^1f(x)dx\Big)\Big(\int_0^1 g(x)dx\Big)$ ?

$\endgroup$

closed as off-topic by choco_addicted, Arnaud D., Jyrki Lahtonen, Jendrik Stelzner, user99914 Aug 14 '18 at 18:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Arnaud D., Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ what mean $f\in\mathcal R[0,1]$ ? $\endgroup$ – Surb Nov 15 '15 at 9:18
  • $\begingroup$ I think is Riemann integrable on [0,1]. Something looks Really weird in this problem... Since $g$ is continuous and periodic on $\mathbb R$, there is a $M$ s.t. $|g(x)|\leq M$ for all $x$. In particular, $|f(x)g(nx)|\leq M|f(x)|\in L(0,1)$, and thus we can theoretically permute limite and integral (dominated convergence theorem), but $\lim_{n\to\infty }f(x)g(nx)$ do not exist since $g$ is periodic... or it exist if $g$ is constant or $f\equiv 0$. Therefore something looks very strange. $\endgroup$ – Rick Nov 15 '15 at 9:28
  • $\begingroup$ @Rick : You are missing one crucial point , you cannot use Dominated Convergence theorem here , the sequence of functions $\{f(x)g(nx)\}$ is not known to be pointwise convergent ! $\endgroup$ – user228168 Nov 15 '15 at 12:16
  • 2
    $\begingroup$ Possible duplicate of integral involving a periodic function $\endgroup$ – Guy Fsone Jan 16 '18 at 18:23
1
$\begingroup$

Put $\displaystyle u_n=\int_0^1f(x)g(nx)dx$. We have by the change of variable $u=nx$: $$u_n=\int_0^n f(\frac{u}{n})g(u)\frac{du}{n}=\frac{1}{n}\sum_{k=0}^{n-1}\int_k^{k+1}f(\frac{u}{n})g(u)du$$

But as $g$ has period $1$: $$\int_k^{k+1}f(\frac{u}{n})g(u)du=\int_0^1f(\frac{t+k}{n})g(t+k)dt=\int_0^1f(\frac{t+k}{n})g(t)dt$$ Put $\displaystyle T_n(t)=\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{t+k}{n})$. We have hence $\displaystyle u_n=\int_0^1 T_n(t)g(t)dt$. Now $T_n(t)$ is a Riemann sum for $f$ (On $\displaystyle I_k=[\frac{k}{n},\frac{k+1}{n}]$, we have $ {\rm Inf}_{u\in I_k}f(u)\leq f(\frac{t+k}{n})\leq {\rm Max}_{u\in I_k}(f(u))$). Hence $\displaystyle T_n(t)\to L=\int_0^1 f(t)dt$. Now there exists $M$ such that $|f(u)|\leq M$ for all $u$, hence we get $\displaystyle |T_n(t)g(t)|\leq M|g(t)|$, and by the convergence dominated theorem, we get $\displaystyle u_n\to \int_0^1Lg(t)dt$, and we are done.

$\endgroup$
  • $\begingroup$ "Now there exists $M$ such that $|f(u)|\leq M$ for all $u$..." Is it true that each Riemann integrable function is bounded? Just asking. $\endgroup$ – Olivier Oloa Nov 15 '15 at 9:31
  • 2
    $\begingroup$ I think that Riemann integrability on a compact interval $[a,b]$ is only for bounded functions (if not, we cannot define upper and lower Riemann sums) $\endgroup$ – Kelenner Nov 15 '15 at 9:34
  • $\begingroup$ Yes, I think you may rather consider lower and upper Riemann sums (a counter example is $f(x)=x^{-1/2}$ over $(0,1], f(0)=0$ which is not bounded). $\endgroup$ – Olivier Oloa Nov 15 '15 at 9:48
  • 1
    $\begingroup$ But your function $x^{-1/2}$ is not Riemann-integrable on $[0,1]$ (It is only Riemann-integrable in the generalized meaning). $\endgroup$ – Kelenner Nov 15 '15 at 9:50
  • $\begingroup$ @Kelenner : Neat solution .. but I was wondering , can we avoid Dominated Convergence theorem ? $\endgroup$ – user228168 Nov 15 '15 at 12:18