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

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


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  • $\begingroup$ 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
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    $\begingroup$ Possible duplicate of integral involving a periodic function $\endgroup$ – Guy Fsone Jan 16 '18 at 18:23

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.

  • $\begingroup$ "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
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    $\begingroup$ 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
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    $\begingroup$ 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