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Prove the following statement or find a counter example:Every convergent sequence that is uniformly bounded on a compact set of real numbers contains a uniformly convergent subsequence.

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Does "uniformly convergent" mean anything other than "convergent", with regard to a sequence? – Dylan Moreland Feb 18 '12 at 7:12
I am assuming he is referring to a sequence of functions on a compact set. – zrbecker Feb 18 '12 at 7:30
@Danikar That makes sense. I mostly wanted the OP to clarify the question. I don't think there's a reason to be vague :) – Dylan Moreland Feb 18 '12 at 20:50
up vote 2 down vote accepted

Define $f_n: [0, 1] \to \mathbb{R}$ by $$f_n(x) = \frac{x^2}{x^2 + (1 - nx)^2}.$$

$|f_n(x)| \le 1$, so uniformly bounded. And $f_n \to f = 0$. But for any subsequence $\{f_{n_k}\}$ we have $f_{n_k}(1/n_k) = 1$, so no subsequence converges uniformly.

See Rudin, Principles of Mathematical Analysis, Chapter 7 page 156.

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Consider the sequence of functions $$f_n(x)=x^n$$ on the interval $[0,1]$. For $x\in[0,1)$, we have $f_n(x)\rightarrow0$ and for $x=1$, we have $f_n(x)\rightarrow 1$. The limiting function of this sequence is not even continuous!

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