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The following problem is from Munkres's Topology (Exercise 6 of Section 21 "The Metric Topology (continued)", 2nd edition).

Exercise: Define $f_n : [0,1] \to \mathbb{R}$ by the equation $f_n(x) = x^n$. Show that the sequence $(f_n(x))$ converges for each $x \in [0,1]$, but that the sequence $(f_n)$ does not converge uniformly.

I can show that for each $x \in [0,1)$, the sequence $(f_n(x))$ converges to $0$ and $f_n(1)$ converge to $1$. However, I failed to show the second part. More generally, I am not sure about the basic idea of how to show that some function sequence does not converge uniformly.

I think the difficulties lie in the quantifiers ($\forall, \exists$) in the definition of uniform convergence.

Uniform Convergence: Let $f_n : X \to Y$ be a sequence of functions from the set $X$ to the metric space $Y$. Let $d$ be the metric for $Y$. We say that the sequence $(f_n)$ converges uniformly to the function $f : X \to Y$ if given [$\forall$] $\epsilon > 0$, there exists [$\exists$] an integer $N$ such that $$d(f_n(x), f(x)) < \epsilon$$ for all [$\forall$] $n \ge N$ and all [$\forall$] $x$ in $X$.

Logically, there seems to be more than one ways to negate the conditions of the uniformality of convergence, by negating some of these four quantifiers. However, I am stuck with them. Therefore,

Problems:
(1) How to show that the function sequence in this Exercise does not converge uniformly?
(2) What are the typical approaches to showing that some function sequence does not converge uniformly?

hengxin
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1 Answers1

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There is a theorem

A uniformly convergent sequence of continuous functions converges to a continuous limit.

Therefore one of the easiest ways to show that a sequence of functions is not converging uniformly is if all the terms are continuous, but the limit is discontinuous. Note that this is the case in your example.

There are times when you may need to show it more directly. For instance, at this point in Munkres's text you may not have the above theorem. What you then need to do is to find a counterexample to the criterion of being uniformly convergent. A good place to start is to visualize the situation. Look at the graph of the limit function (assuming you're in one variable). Uniform convergence means that eventually all the terms of the sequence have graphs which lie within a "epsilon-tube" of the graph of the limit function. So if they are not converging uniformly, you have to find out why the graphs of some terms will always stray outside an epsilon-tube. More precisely, you have to find some $\epsilon$ such that there are arbitrarily high $n$ where $\sup_{x\in D} |f_n(x)-f(x)|\geq \epsilon$.

In this case, if you do the visualization, you find out that the problem is near $x=1$. Note that for all $n$ there will be some point such that $x^n=1/2$. It will be a distance of $1/2$ from the limit function, so choose $\epsilon = 1/4$, and you're home!

P.S. There is a systematic way to negate quantifiers, and you should surely familiarize yourself with it. Search for a link, or try this http://www.math.cornell.edu/~hubbard/negation.pdf

Eric Auld
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