4
$\begingroup$

I am new to this type of mathjax formatting, so feel free to refine this question. I've tried my best with the formatting.

Anyway, this is a question I've encountered in a homework sheet, and I do not even know where to begin:

Let $f:[0, \infty) \rightarrow \mathbb R$ be a continuous real valued function such that $f(x+1)=f(x)$ for all $x\ge0$. If $g:[0,1] \rightarrow \mathbb R$ is an arbitrary continuous function, show that $$\lim_{n\to \infty}\int_0^1g(x)f(nx)\,dx = \left(\int_0^1g(x)\,dx\right)\left(\int_0^1f(x)\,dx\right).$$

We were given a hint:

$$\int_0^1 g(x)f(nx)\,dx = \frac{1}{n} \sum_{i=1}^n \int_{i-1}^i g\left(\frac{u}{n}\right)f(u)\,du,$$ and put $t = u - i + 1.$

I have absolutely no idea where to begin. My thoughts are that this question will involve the use of Lebesgue's convergence theorem, and perhaps the monotone convergence theorems. I have a very basic understanding of these theorems but still struggle when they need to be applied. My understanding is that:

  1. Function must be Riemann Integrable
  2. $f_n \rightarrow f$ almost everywhere
  3. $|f_n| \le g \in L^1$

I understand 1., and kind of understand 3. but I'm never able to prove 2. without any help. In fact, I only have a vague understanding of 2. and 3.

I've attempted many questions, but just cannot finish one without help. My take is that I lack basic understanding on measure theory and I also lack practice, although I've been spending a large portion of time on this subject this semester. Everything is new and extremely difficult. I'd appreciate some books on the subjects, the lecture notes are really good, but I don't think it's enough at this point. I want something better than a mere pass (and if I aim for a pass, I will fail the subject) -- I want to actually understand it.

Any help and recommendations are appreciated. Thanks!

$\endgroup$
  • $\begingroup$ I've edited this multiple times for all the functions to show up, new to this. $\endgroup$ – Hypergeometry May 12 '18 at 16:03
  • $\begingroup$ I have edited as best as I can. But, as written, the question makes no sense because there is no information about $f_n$. $\endgroup$ – Martin Argerami May 12 '18 at 16:09
  • $\begingroup$ My bad, it's supposed to be $f(nx)$. I'll edit the question to fix this. $\endgroup$ – Hypergeometry May 12 '18 at 16:11
  • $\begingroup$ Yes, I just noticed. Thanks for editing. $\endgroup$ – Martin Argerami May 12 '18 at 16:12
  • $\begingroup$ If you know the RIemann-Lebesgue lemma then you may notice that this result is very much in the same stream if you set $\phi_n(x)=f(nx)-\int_0^1f(x)\,dx$ (as they have zero average over the period as $\sin$, $\cos$ do). The similar to R-L approach would work here: prove first for simple functions $g$ and then approximate a general $g$ uniformly by simple ones. Does it make sense? $\endgroup$ – A.Γ. May 12 '18 at 18:07
4
$\begingroup$

Proof. Let us only assume that $f$ is Lebesgue integrable on $[0, 1]$ and $g$ is Riemann integrable on $[0, 1]$, as this generalization rarely harm the essence of the argument. Then

$$ \int_{0}^{1} f(nx)g(x) \, dx \stackrel{nx=u}{=} \frac{1}{n}\int_{0}^{n} f(u)g\left(\frac{u}{n}\right)\,du = \frac{1}{n}\sum_{i=1}^{n} \int_{i-1}^{i} f(u)g\left(\frac{u}{n}\right)\,du. $$

Substituting $v = u-i+1$, we get

$$ \int_{0}^{1} f(nx)g(x) \, dx = \int_{0}^{1} f(v) \left[ \frac{1}{n}\sum_{i=1}^{n} g\left(\frac{v+i-1}{n}\right) \right] \,dv. $$

Now write $\Pi_n = \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\}$ and let $U_n = U(g, \Pi_n)$ and $L_n = L(g, \Pi_n)$ denote the upper Riemann sum and the lower Riemann sum for $\Pi_n$, respectively. Then $U_n$ and $L_n$ converge to the integral $I := \int_{0}^{1} g(x) \, dx$ as $n\to \infty$. Moreover,

$$ \forall v \in [0, 1] \ : \quad L_n \leq \frac{1}{n}\sum_{i=1}^{n} g\left(\frac{v+i-1}{n}\right) \leq U_n . $$

Of course we also have $L_n \leq I \leq U_n$. So

$$ \left| \frac{1}{n}\sum_{i=1}^{n} g\left(\frac{v+i-1}{n}\right) - I \right| \leq U_n - L_n $$

and therefore

$$ \left| \int_{0}^{1} f(nx)g(x) \, dx - I \int_{0}^{1} f(v) \, dv \right| \leq (U_n - L_n) \int_{0}^{1} |f(v)| \, dv \xrightarrow[n\to\infty]{} 0. $$


Idea. Loosely speaking, $g$ hardly changes its value while $f(nx)$ completes one period on an interval of length $\frac{1}{n}$. In this way, behaviors of $g(x)$ and $f(nx)$ are almost decoupled for large $n$. ($g(x)$ can only see the 'averaged value of $f(nx)$', while $f(nx)$ can hardly detect the change of values of $g(x)$.) The above argument provides a quantitative version of this intuition.

$\endgroup$
  • $\begingroup$ Disregard the previous comment. $\endgroup$ – Mark Viola May 12 '18 at 17:29
  • $\begingroup$ Thanks! This makes a lot more sense to me than invoking epsilon, although obviously I need to revise that. My question however remains, the differential after substitution is $x=u/n$. 1/n is simply brought to the front and the boundaries are changed? Is that always true? It's almost 4am so I better sleep, I'll look into it deeper later in the afternoon. $\endgroup$ – Hypergeometry May 12 '18 at 17:48
  • $\begingroup$ I think I get it? It's just the product rule... silly me. $\endgroup$ – Hypergeometry May 12 '18 at 17:51
2
$\begingroup$

With the substitution $v=nx$, you get $$ \int_0^1 g(x)f(nx)\,dx=\frac1n\,\int_0^n g(v/n)\, f(v)\,dv=\frac1n\,\sum_{k=1}^n\int_{k-1}^k g(v/n)\,f(v)\,dv. $$ Fix $\varepsilon>0$. Since $g$ is continuous on $[0,1]$, it is uniformly continuous by compactness, so there exists $\delta>0$ such that $|g(y)-g(x)|<\varepsilon$ whenever $|x-y|<\delta$. So, if $n>1/\delta$, then $$|g(v_1/n)-g(v_2/n)|<\varepsilon$$ for all $v_1,v_2\in[k,k+1]$. So \begin{align} \left|\int_0^1 g(x)f(nx)\,dx-\frac1n\,\sum_{k=1}^n\int_{k-1}^k g((k-1)/n)\,f(v)\,dv\right|<\varepsilon. \end{align} Now, since $f(x+1)=f(x)$, \begin{align} \frac1n\,\sum_{k=1}^n\int_{k-1}^k g((k-1)/n)\,f(v)\,dv &=\frac1n\,\sum_{k=1}^ng((k-1)/n)\int_{k-1}^k \,f(v)\,dv\\ \ \\ &=\frac1n\,\sum_{k=1}^ng((k-1)/n)\int_{0}^1 \,f(v)\,dv. \end{align} If we choose $n$ big enough, we can get $\frac1n\,\sum_{k=1}^ng((k-1)/n)$ arbitrarily close to $\int_0^1 g(x)\,dx$. That is, we may choose $n$ big enough so that $$ \left|\int_0^1 g(x)f(nx)\,dx-\int_0^1 g(x)\,dx\int_0^1 f(x)\,dx\right|<2\varepsilon. $$

$\endgroup$
  • $\begingroup$ Thanks for the response! When $v=nx$, the integral becomes $\int_0^1g(v/n)f(v)\,d(v/n)$. So 1/n can just be brought outside as a multiplier, and the boundaries of $(0,1)$ swapped to $(n,0)$? I also don't understand how $u=v-k+1$ was used, although I can see that integrating from 0 to n would just be summing up those segments which are illustrated in the summation. The rest would require some studying, I haven't dealt with hard math like this in 2 years; would probably be a good idea to revise some notes. I kind of understand what's going on. $\endgroup$ – Hypergeometry May 12 '18 at 17:41
  • $\begingroup$ The $u$ is to show that $\int_{k-1}^k f(v)\,dv=\int_0^1 f(u)\,du$. $\endgroup$ – Martin Argerami May 12 '18 at 18:48
0
$\begingroup$

This is essentially the same answer as the others, except that it focuses on the fact that a particular Riemann approximation converges to $\int g$ almost everywhere (ae.) after which we can use the dominated convergence theorem.

If $g$ is Riemann integrable, then for ae. $t$, the function $g_n(t) = {1 \over n} \sum_{k=0}^{n-1} g({t+k \over n})$ satisfies $L(g,P_n) \le g_n(t) \le U(g,P_n)$, where $P_n$ is the partition $(0,{1 \over n}, \cdots, {2 \over n}, \cdots, { n \over n})$.

Since $g$ is Riemann integrable, by choosing $n$ large enough, we see that $g_n(t) \to \int g$ for ae. $t$.

Since $\int_0^1 g(x) f(nx) dx = \int g_n(t) f(t) dt$, we can apply the dominated convergence theorem to get the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.