# Show that $x_n \rightarrow 0$

Let $f:[0,1] \rightarrow \mathbb{R}$ continuous, such that $f(0)=0$

We set $x_n=\int_0^1{f(x^n)}dx$

Show that $x_n \rightarrow 0$

 The function $f$ is continuous at a closed interval $\Rightarrow$ $f$ is bounded $\Rightarrow \exists M>0: |f(x)| \leq M, \forall x \in [0,1]$

The function $f$ is continuous and $f(0)=0$ $\Rightarrow \lim_{x \rightarrow 0}{f(x)}=0 \Rightarrow \forall \epsilon >0 \text{ } \exists \delta >0: \forall x \in [0,1], |x-0| < \delta \Rightarrow |f(x)-f(0)|< \epsilon$

So $0 \leq x< \delta \Rightarrow |f(x)| < \epsilon$  How can I continue?

• You are interested in $x$ not near $0$, but in the two intervals $[0,\epsilon^\ast]$ and $[\epsilon^\ast,1]$, where $\epsilon^\ast$ is "small." Show that for any $\epsilon^\ast\gt 0$, there is an $N$ such that if $n\gt N$ and $x\lt 1-\epsilon^\ast$, then $x^n\lt \epsilon^\ast$. Apr 21, 2014 at 16:12
• Because $1^n=1$. Apr 21, 2014 at 16:14
• The answer by Hagen von Eitzen goes along these lines. Sure, for any $\epsilon^\ast$ there is an $N$ such that if $n\gt N$ and $0\lt x\lt 1-\epsilon^\ast$, we have $|x^n|\lt \epsilon^\ast$. I think that for this problem that can be taken as obvious. You can find an explicit $N$ by using logarithms. Apr 21, 2014 at 16:41
• Yes, sorry about the typo, it is $[0,1-\epsilon^\ast]$ and $[1-\epsilon^\ast,1]$. Apr 21, 2014 at 16:43
• You are welcome. The basic idea is geometric, $x^n$ will be tiny except when we are very close to $1$. But the integral from $1-\epsilon^\ast$ to $1$ is $\lt M\epsilon^\ast$. However, writing out the details takes some concentration. Apr 21, 2014 at 17:09

Given $\epsilon>0$, there exists $\delta>0$ such that $|f(x)|<\frac12\epsilon$ for $0\le x<\delta$. Also, there exists $q\in(0,1)$ such $\frac\epsilon2\cdot q+(1-q)\cdot M<\epsilon$. Finally, there exists $N$ such that $q^N<\delta$. Then for all $n\ge N$ $$\left|\int_0^1f(x^n)\,\mathrm dx\right|\le \left|\int_0^qf(x^n)\,\mathrm dx\right|+\left|\int_q^1f(x^n)\,\mathrm dx\right|\le q\cdot\epsilon2+(1-q)M<\epsilon$$

• Ok!!! Thanks a lot!!! Apr 21, 2014 at 16:55

Hints using the notation and stuff you already did:

For all $\;\epsilon>0\;$ there exists $\;K\in\Bbb N\;\;s.t.\;\;n>K\implies |f(x^n)|<\epsilon\;$ (why?) , so:

$$\left|\int\limits_0^1f(x^n)\,dx\right|\le\int\limits_0^1|f(x^n)|dx\le\int\limits_0^1\epsilon\,dx=\epsilon$$

• Could you give me a hint why $|f(x^n)|< \epsilon$ ?? I got stuck.. Apr 21, 2014 at 16:12
• @MaryStar: $$x\neq1\,,\,x^n\xrightarrow[n\to\infty]{}0\;\;\text{(why?)}\; \implies f(x^n)\xrightarrow[n\to\infty]{}f(0)\;\;\text{(again, why?)}$$ Apr 21, 2014 at 16:14
• The answer of the first question is: Since $x<1 \Rightarrow \lim_{n \rightarrow \infty}{x^n}=0$. For the second question the answer is the continuity of $f$. Is this correct? Apr 21, 2014 at 16:18
• Indeed so, @MaryStar...perhaps you shall need to add a word about the special point $\;x=1\;$, but as the value of the integral doesn't change if we modify a finite number of the function's values... Apr 21, 2014 at 16:23
• I understand!!! Thank a lot!!! Apr 21, 2014 at 16:56