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I've gotten the guess from the two links.

links) integral involving a periodic function & Prove $\lim_{n\to \infty}\int_0^Tg(x)h(nx) dx=\cfrac{1}{T}(\int_0^Tg(x)dx)(\int_0^Th(x)dx)$

  • For the periodic function $g$ whose period is $p$. Here the $f$ and $g$ is continuous on $[a,b]$.

    $\lim\limits_{n \to \infty} \int_a ^{b} f(x) \cdot g(nx) dx = (\int_a^{b}f(x)dx)(\frac{1}{p}\int_a ^{b} g(x) dx) $

As you can see, I expand this $[0,1]$(or $[0,T]$) to $[a,b]$ for any real numbers $a, b(>a)$. I tried to prove my guess but Whenever I tried it, I failed. The bottom line is I eager to know prove method or counterexamples for my guess.

Regards.

kechang lee
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  • The right conjecture would presumably have the factor of $\frac1{b-a}$, not $\frac1p$ (consider for example if $f$ and $g$ are constant). – Greg Martin Jul 21 '22 at 03:29
  • Both linked questions assume both functions are continuous. Do you want to require that? We might get away with the weaker condition that they're integrable and bounded. But without some such restriction, there are functions like $f(x) = g(x) = (1-x+\lfloor x \rfloor)^{-1/2}$, integrable on any interval but $\int_0^b f(x) g(nx), dx = \infty$ – aschepler Jul 21 '22 at 11:49
  • @aschepler oh yes. I edited as the continuity. – kechang lee Jul 21 '22 at 12:02

3 Answers3

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In the formula you quote, $T$ is the period of $h$ and you integrate over $[0,T]$, one period of the function. There is no reason to think we get any nice formula integrating on $[a,b]$, unless the period is $b-a$.

example
$f(x) = 1, g(x) = \cos x$. Then $$ \lim_{n\to\infty}\int_a^b f(x)g(nx)\;dx =\lim_{n\to \infty}\frac{\sin(nb)-\sin(na)}{n} = 0 \tag1$$ but $$ \left(\int_a^b f(x)\;dx\right)\left( \int_a^b g(x)\;dx\right) = (b-a)\big(\sin(b)-\sin(a)\big) \tag2$$
Note $(2)$ is not $0$ unless $\sin(a) = \sin(b)$

Remark. For your limit $$ \lim_{n\to\infty}\int_a^b f(x)g(nx)\;dx $$ see: the Riemann-Lebesgue lemma

GEdgar
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I believe that the equality is not true. I invoke the Riemann-Lebesgue Lemma which says that :

$\displaystyle \lim_{n \to +\infty}\int_{a}^{b}f(x)sin(nx)dx=0$. (For a continuous $f$ on $[a,b]$)

So, if we set $g(y)=siny\,\,$ which has period $2\pi$ we have the l.h.s. equal to zero, and the r.h.s. equal to :

$\int_{a}^{b}f(x)dx$ $\dfrac{1}{2\pi}\int_{a}^{b}sinxdx=$$\int_{a}^{b}f(x)dx\,\dfrac{-cosb+cosa}{\pi}$

which is NOT necessarily equal to zero.

1

One thing that remained unresolved is whether the equality is true when $a=0$ and $b=T$ the period of the function $h$. I have tried many $h$ of trigonometrical form and the result seems to be correct. But I think the following example disproves even this case.

Let h be defined as follows:

$h(x)=1-(x-1)^{2}$ on $[0,2]$

$h(x)=1-(x-3)^{2}$ on $[2,4]$

$h(x)=1-(x-5)^{2}$ on $[4,6]$ e.t.c.

which has period $T=2$ and the following graph:

enter image description here

Every $nx$ belongs to some $[2n,2n+2]$ where the function $h(x)=1-[x-(2n+1)]^{2}$. Calculating (for $g(x)=x$) the integral:

$\int_{0}^{2}x[1-(nx-(2n+1))^{2}]dx$=$-\dfrac{4}{3}n^{2}-\dfrac{8n}{3}$

which tends to $-\infty$, while the r.h.s. integrals are finite!! Sorry for any errors in the integration!