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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

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

  • what mean $f\in\mathcal R[0,1]$ ? – Surb Nov 15 '15 at 09:18
  • 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. – Rick Nov 15 '15 at 09:28
  • @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 ! –  Nov 15 '15 at 12:16
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    Possible duplicate of integral involving a periodic function – Guy Fsone Jan 16 '18 at 18:23

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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.

Kelenner
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  • "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. – Olivier Oloa Nov 15 '15 at 09:31
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    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) – Kelenner Nov 15 '15 at 09:34
  • 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). – Olivier Oloa Nov 15 '15 at 09:48
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    But your function $x^{-1/2}$ is not Riemann-integrable on $[0,1]$ (It is only Riemann-integrable in the generalized meaning). – Kelenner Nov 15 '15 at 09:50
  • @Kelenner : Neat solution .. but I was wondering , can we avoid Dominated Convergence theorem ? –  Nov 15 '15 at 12:18
  • @SaunDev Sorry, I have left my computer for some times. If you have as hypothesis that $f$ is continuous, then $T_n(f)(t)\to L=\int_0^1f(t)dt$ uniformly on $[0,1]$. To see this, note that $f$ is uniformly continuous, and bound $|f(\frac{t}{n}+\frac{k}{n})-f(k/n)|$ uniformly with respect to $k$ using the fact that $t/n\leq 1/n$. Then you have only to use that $g$ is Riemann-integrable, hence bounded. to show $T_n(t)g(t)\to Lg(t)$ uniformly., and you do not need DCT. – Kelenner Nov 15 '15 at 15:08