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Supose that $f\colon [0,1] \rightarrow \mathbb R$ is a continuous function. Show that: $$\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)$$

I'm studying the Riemann Integral theory and I have not idea to how do this. I need some hint to start because I do not see how to apply the Riemann sum or other theory concept to be able to help me.

Git Gud
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Felipe
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  • The title isn't supposed to be the first line of your question. – Git Gud Jan 04 '14 at 20:32
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    @GitGud Hi, why not ? – Amr Jan 04 '14 at 21:13
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    @Amr The title is supposed to describe the question in short, so people can easily search for it and known what's it is about without having to open it. And I, personally, once I open a question, I don't even read the title anymore. It's in a different format, different color, different everything. – Git Gud Jan 04 '14 at 21:18
  • I'm not sure if you have studied Lebesgue or not. But simplest argument here is that $f$ is bounded, and $f(x^{n})$ converge almost everywhere to a constant function take value $f(0)$, which means it also converge in measure, and thus the integral converge to $\int_{0}^{1}f(0)dx=f(0)$ – Gina Jan 04 '14 at 21:19
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    @GitGud This title lets one know what's inside the question before opening it. – Amr Jan 04 '14 at 21:24
  • @Amr I have no problem with the title being the same as the first line of the question. But at the time of my comment, the question continued from the title. That's what I have a problem with. – Git Gud Jan 04 '14 at 21:26
  • @GitGud Ahhh I see. – Amr Jan 04 '14 at 21:27

2 Answers2

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Since $g(x)=f(x)-f(0)$ is continuous iff $f(x)$ is, and $g(0)=0$, we may assume that $f(0)=0$ in the first place.

Hint: The idea is that if $f$ is continuous at $x=0$ and $f(0)=0$; then $f_n(x)=f(x^n)$ goes to $0$ uniformly over $[0,\eta]$ for any $1>\eta>0$. Thus, given any $\varepsilon>0$; it suffices to take $\eta$ sufficiently close to $1$ to make the integral over $[\eta,1]$ tiny, and use uniform convergence to kill of the integral over $[0,\eta]$.

SPOILERS

Given $\varepsilon>0$, there exists $\delta>0$ such that $0<t<\delta\implies |f(t)|<\varepsilon$.

Let $M>0$ be such that $|f(x)|<M$ over $[0,1]$. Given this $\delta>0$, choose $1>\eta>0$ such that $M(1-\eta)<\varepsilon$, and choose $N$ such that $0\leq x\leq \eta$ implies $0\leq x^N \leq\delta$ (the same will hold true for $n>N$). Then for $n>N$ we will have

$$\left|\int_0^1f(x^n)dx\right|\leqslant \int_0^\eta |f(x^n)|dx+\int_\eta^1 |f(x^n)|dx\leqslant \eta \varepsilon+M(1-\eta)<(1+\eta)\varepsilon<2\varepsilon$$

ADD Note that by our set up, if $x\in[0,\eta]$, then $0\leq x^n\leq \delta$. This means that $|f(x^n)|<\varepsilon$ over $[0,\eta]$, so $$\int_0^\eta |f(x^n)|dx\leqslant \int_0^\eta \varepsilon dx=\eta\varepsilon$$

Pedro
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  • My attempt to proof without supposing that f(0)=0: Given $\varepsilon>0$, there exists $\delta>0$ such that $0<t<\delta \Rightarrow |f(t)-f(0)|<\varepsilon$.Let $M>0$ be such that $|f(x)-f(0)|<M$ over $[0,1]$. Given this $\delta>0$, choose $1>\eta>0$ such that $M(1-\eta)<\varepsilon$, and choose $N$ such that $0\leq x\leq \eta$ implies $0\leq x^N \leq\delta$ (the same will hold true for $n>N$). Then for $n>N$ we will have $$\left|\int_0^1f(x^n)-f(0)dx\right|\leqslant \int_0^\eta |f(x^n)-f(0)|dx+\int_\eta^1 |f(x^n)-f(0)|dx\leqslant \eta \varepsilon+M(1-\eta)<(1+\eta)\varepsilon<2\varepsilon$$ – Felipe Jan 05 '14 at 18:33
  • I know your proof is correct but I made these changes to make sure that I did – Felipe Jan 05 '14 at 18:34
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Here's a slightly alternative way. As already mentioned, if the fact is true for $f(0)=0$ it is true for any $f(0)\neq 0$, so assume $f(0)=0$. Do a change of variables $u=x^n$. Then $du = n x^{n-1} dx = n u^{1-1/n} dx$ and so $$I_n = \int_0^1 \frac{1}{n} u^{1/n-1} f(u) du = \int_0^\epsilon \frac{1}{n} u^{1/n-1} f(u) du + \int_\epsilon^1 \frac{1}{n} u^{1/n-1} f(u) du.$$

The first term is bounded in absolute value by $$\sup_{u \in [0, \epsilon]} |f(u)| \cdot \int_0^\epsilon \frac{1}{n} u^{1/n-1} du=\sup_{u \in [0, \epsilon]} |f(u)| \;\epsilon^{1/n}. $$ For the second integral, $u^{1/n-1}\leq \epsilon^{1/n-1}$ on our region of integration, so the second integral is bounded in absolute value by $$\frac{\epsilon^{1/n-1}}{n} \int_\epsilon^1 |f(u)| du.$$

Using these bounds, $$\limsup_{n \to \infty} |I_n| \leq \limsup_{n \to \infty} \left[ \sup_{u \in [0, \epsilon]} |f(u)| \;\epsilon^{1/n} + \frac{\epsilon^{1/n-1}}{n} \int_\epsilon^1 |f(u)| du. \right] =\sup_{u \in [0, \epsilon]} |f(u)|+0.$$ By the continuity of $f$, we have $\sup_{u \in [0, \epsilon]} \to 0$ as $\epsilon \to 0$, so $\limsup_{n \to 0} |I_n| = 0$ and hence $\lim_{n \to 0} I_n = 0$.

I'm pretty sure this way you could beat this problem into the form of a mollifier, that is an approximation to identity, if you wanted.

abnry
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