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For the following sequence of functions and its limit function, we can see that $f_n(x)$ is clearly pointwise convergent

$$f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\ f(x) = \begin{cases}0&\text{if } x\in[0,1)\\1&\text{if } x=1\end{cases}$$

However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is:

$$|f_n(x) - f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N$$

Now most sources present an argument along the lines of: assume that $f_n(x)$ is uniformly convergent and that $0 < x < 1$, this means that $x^n<\epsilon$ whenever $n>N$. Specifically, this would mean $x^{N+1}<\epsilon$ for some fixed $N$. But if we now pick $x$ such that $1 > x > ε^{\frac{1}{N+1}}$, then this would lead to a contradiction, therefore $f_n(x)$ is not uniformly convergent.

However, I was wondering why couldn't we take $n$ to infinity. If $0 < x < 1$, then $\lim_{n\rightarrow \infty} |f_n(x) - f(x)| = 0$ (which is less than $\epsilon$). Now since $|f_n(x) - f(x)|$ will always be $0$ if $n > \infty$, then wouldn't this be uniformly convergent?

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I think that you hint would prove that $\{f_n\}$ converges uniformly to zero on every interval $[0,b]$ with $0<b<1$. – Siminore Aug 23 '12 at 14:32
up vote 8 down vote accepted

You can't just take $n=\infty$. In effect what you have done is to verify pointwise convergence, not uniform convergence.

Uniform convergence fails because, for every $n$, $\sup_{x\in[0,1]} |f_n(x)-f(x)| = 1$, which you can see by continuity.

Alternatively, uniform convergence must fail because each $f_n$ is continuous, $f$ is not continuous, and a uniform limit of continuous functions is continuous.

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Hmm ok but why can't we just take n to infinity? The definition for uniform convergence I am using: $$|f_n(x) - f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N$$ does not preclude n being taken to infinity does it? – BYS2 Aug 25 '12 at 3:36
@BYS2 You must remember the quantification, especially the order of quantification. The definition is this: $f_n\to f$ uniformly if for every $\epsilon>0$ there exists $N$ (which can only depend on $\epsilon$) such that for all $n\geq N$ and all $x \in [0,1]$ we have $$|f_n(x) - f(x)| < \epsilon.$$ An exactly equivalent way of saying this is $$\sup_{x\in[0,1]} |f_n(x) - f(x)| \to 0 \text{ as $n\to\infty$}.$$ – Sean Eberhard Aug 25 '12 at 6:36
Ahh ok I see... its just a switching of the order of quantification.. thanks! – BYS2 Aug 26 '12 at 4:35

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