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I am currently learning topology from Munkres. The question below is an exercise in section 45.

Let $f_n\colon I\to \mathbb{R}$ be the function $f_n(x)=x^n$. The collection $F=\{f_n\}$ is pointwise bounded but the sequence $(f_n)$ has no uniformly convergent subsequence; at what point or points does $F$ fail to be equicontinuous.

Any help would be appreciated. Thank You.

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1  
Can you find any point in $I$ where the family of functions is equicontinuous? –  you Apr 13 '12 at 0:31
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It might help to think about what the limit function is. –  Tyler Apr 13 '12 at 0:32
    
hint: consider the point on the right hand side of the interval $I$ (1). Can you find the open set required by equicontinuity for an $\epsilon<1$ for ALL functions in $F$? –  Ralth Apr 13 '12 at 0:34

1 Answer 1

Since for $|c|<1$, and $|x|\leq c|$, $|f'_n(x)|=n|x^{n-1}|\leq n |c|^{n-1}$ and the sequence $\{n c^{n-1}\}$ is bounded, $\{f_n\}$ is equi-continuous at each point of $(-1,1)$.

We don't have equi-continuity at $1$, otherwise we would be able to find $\delta>0$ such that for each $n$, $|1-x^n|\leq \frac 12$ for $|1-x|\leq \delta$. THis gives $x^n\geq \frac 12$ for all $n$, which is not possible. A similar argument applies for $-1$.

We don't have equi-continuity at $x_0$ where $|x_0|>1$. Otherwise $|x^n-x_0^n|\leq 1$ for all $n$ and $|x-x_0|\leq\delta$ for some $\delta$, hence taking $x_n=x_0+1/n$ for a $n$ large enough, $$\left|\left(x_0+\frac 1n\right)^n-x_0^n\right|\leq 1$$ hence $$|x_0|^n\left|\left(1+\frac 1{n|x_0|}\right)^n-1\right|\leq 1.$$ As $\left|\left(1+\frac 1{n|x_0|}\right)^n-1\right|\to |e^{1/|x_0|}-1|\neq 0$, we get a contradiction.

Conclusion: $\{f_n\}$ is equi-continuous on the points of $(-1,1)$, and fail in all the others.

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