I need to prove the following:

Suppose $M$ a metric space and $N$ a complete metric space. Show that if a sequence of continuous functions $f_n:M\to N$ converges uniformly in a subset $X$, then $f_n$ converges uniformly in $\overline {X}$. (where $\overline{X}$ is the closure of $X$).

Well, first of all, what excatly does it means to say that a sequence of functions converges uniformly to $f$? I understand it by: for all $\epsilon>0$ there exists $n_0$ such that

$$n>n_0\implies |f_n(x)-f(x)|<\epsilon$$

for all $x\in X$, and all $n>n_0$.

Question: is this $n_0$ dependent of $\epsilon$ only?

Also, how do I relate this result to the closure of $X$?

I know of this theorem: suppose $f:A\to N$, if there exists $b=\lim_{x\to a} f(x)$, we have $b\in \overline{f(A)}$. Is this helpful?

  • $\begingroup$ To your first question: yes, $n_0$ should depend on $\varepsilon$ only, in particular it must not depend on the choice of $x\in X$. This is what 'uniform' refers to. $\endgroup$
    – Thomas
    Jun 5 '16 at 5:27

Yes, the $n_0$ only depends on $\varepsilon$, that is why it's called "uniform" (i.e. for all $x$ in $X$ at the same time!) convergence. If the $n_0$ is allowed to vary between $x$, it's called pointwise convergence.

You also omitted the statement "there exists $f: X \rightarrow N$ such that ..." and you probably know that a uniform limit of continuous functions is continuous, so $f$ is continuous on $X$, which might come in useful.

Now pick $p \in \overline{X}$. First you have to figure out what $f(p)$ should be. Indeed we can find sequences $(x_n)$ from $X$, that converge to $p$. And you would hope that $f(x_n) \rightarrow f(p)$: if the result would hold the supposed $f$ on $\overline{X}$ would also be continuous, if convergence were uniform.... But does the limit exist? Does it depend on the chosen sequence?

Now is the time to use that $N$ is a complete metric space. Show that $f(x_n)$ is Cauchy in $N$ for every choice of sequence converging to $p$. Use unicity of limits in metric spaces to show that the limit does not actually depend on the specific sequence. Having done that, you have your candidate limit function on $\overline{X}$.


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