Recall Dini's Theorem:

Let $K$ be a compact metric space. Let $f:K\to \mathbb{R}$ be a continuous function and $f_{n}:K\to \mathbb{R}$, $n \in \mathbb{N}$ be a sequence of continuous functions. If $\{f_{n}\}_{n \in \mathbb{N}}$ converges to $f$ and if $f_{n}(x)≥f_{n+1}(x)$ for all $x\in K$ and all $n \in \mathbb{N}$, then $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $f$.


Take $f_{n}:[0,\frac{1}{2}] \to \mathbb{R}$ given by $f_{n}(x) = x^{n}$. We have that $[0,\frac{1}{2}]$ is a compact set and $\{f_{n}\}_{n \in \mathbb{N}}$ converges to $f = 0$, where $f_{n}'s$ and $f$ are continous. But $\{f_{n}\}_{n \in \mathbb{N}}$ doesn't converge uniformly to $f$, (suppose that $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $f=0$ and take $\epsilon = \frac{1}{4}$ so there exist $n_{0} >0$ such that to $n > n_{0}$ imply that $|x^{n}| < \frac{1}{4}$. Take $x = (\frac{1}{3})^{\frac{1}{n}}$ so $\frac{1}{3} < \frac{1}{4}$ !!).

Where am I missing?

Thank you


I assume you meant: take $x =\left( \frac1{3}\right)^{\frac1{n_0 + 1}}$ or something (you have to talk about some fixed $n$).

The problem is that as soon as $n \ge 3$, we have $3/2^n \le 3/8 < 1$, so $1/3^{1/n} > 1/2$ and you are no more in $[0,1/2]$.


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