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Let $f_n$ be a sequence of functions defined on $[a,b]$. Suppose that for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly. Prove that $f_n$ converges uniformly on $[a,b]$

I know that since for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly, meaning for all $\epsilon >0$ there exist an $N>0$ such that $n>N$ implies $|f_n(c)-f(c)|<\epsilon$.

I know that $c$ is arbitray any where in $[a,b]$, but I don't know how to argue that $f_n$ converges uniformly on $[a,b]$ formally.

ok, here is what I got

Assume that $f_n$ be a sequence of functions defined on $[a,b]$. Suppose that for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly. Since $[a,b]$ is compact, there exists finitely many number of point $c_1,c_2,c_3, ... \in [a,b]$ such that

$[a,b] \subset \cup _{k=1}^n (c_k - \frac {\delta(c_k)}{2},c_k + \frac {\delta(c_k)}{2})$

since or every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly, meaning for all $\epsilon >0$ there exist finitely number of $Ns>0$ such that $n>Ns$ implies $|f_n(c)-f(c)|<\epsilon$. Hence, $f_n$ converges uniformly on $[a,b]$?

Davide Giraudo
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Diane Vanderwaif
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  • Use Borel's lemma, which is the following: if you have some family of the open intervals, covering the closed interval $[a,b]$, then from that family of intervals you can choose finitly many intervals also covering $[a,b]$ – Mher Mar 08 '14 at 13:31

2 Answers2

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Your first instinct should be, "$[a,b]$ is compact so of course sequences of functions behave nicely." ;) A general theme is that things are nice on compact spaces, so we'll want to use the compactness of $[a,b]$ to get this result.

You have a small interval about each $c\in[a,b]$ in which $f_n$ converges uniformly. This gives you an open cover of $[a,b]$, which by compactness has a finite subcover. Now (loosely speaking) for any $\epsilon > 0$ you have a finite number of $N$s beyond which $|f_n - f| < \epsilon$.

More precisely, given any $\epsilon > 0$, what you need is an $N$ so that for any $x\in[a,b]$, if $n>N$, then $|f_n(x) - f(x)| < \epsilon$. You know this is true for an interval $U_c$ about $c\in[a,b]$: given $\epsilon>0$, there is an $N_c$ such that for each $x\in U_c$ $|f_n(x) - f(x)|<\epsilon$.

The $U_c$ form a cover of $[a,b]$, which by compactness has a finite subcover $U_{c_i}$ (where $i$ runs through a finite indexing set). Given $\epsilon>0$, these have $N_{c_i}$ such that if $n>N_{c_i}$ and $x\in U_{c_i}$, $|f_n(x) - f(x)|<\epsilon$.

How should you choose a single $N$ for all of $[a,b]$, given that you know the $N_{c_i}$ for $U_{c_i}$ and that $[a,b]\subseteq \cup_i U_{c_i}$?

Neal
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  • I updated my attempt, would you check it for me? I'm still not sure that I understand this correctly, since this chapter was self study, so I'm confused a lot. – Diane Vanderwaif Mar 08 '14 at 13:50
  • @DianeVanderwaif Your confusion appears to be in picking $N$ from the subintervals, so I've updated my answer accordingly. – Neal Mar 08 '14 at 14:42
  • should I choose $N = min(N_{c_1}, N_{c_2},...)$? Because if it's the smallest then it guarantee that the implication will hold, am I correct? – Diane Vanderwaif Mar 08 '14 at 14:53
  • @DianeVanderwaif Not quite ... if you take the smallest one, then near a different $c_j$ you could have $n < N_{c_j}$ but still bigger than $N$. – Neal Mar 08 '14 at 14:57
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    oh, so if I take $N$ to be the biggest one then I'm guaranteed that if $n>N$ then $n$ is greater than any of those, and the conclusion is hold of all $n$, yes? – Diane Vanderwaif Mar 08 '14 at 15:02
  • @DianeVanderwaif That sounds right. Be sure to verify for yourself that the logic holds! – Neal Mar 08 '14 at 19:00
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You can use a trick similar to mathematical induction but on real line:

Let $I = \{ x \in [a,b]: f_n \mbox{ converges uniformly on } [a,x] \}$

We know:

For every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly

Take $c=a$ in the above statement, and then there exists an interval near $a$ that $f_n$ converges uniformly. Therefore $I$ is nonempty, $\sup(I)$ exists and let $S = \sup(I)$

We want to prove $S = b$, then we are done.

Suppose $S < b$,

$f_n$ converges uniformly on $I$ by definition of $I$

Take $c=S$, then there exists an interval $J$ near $S$ that $f_n$ converges uniformly. Then, take a subset $[S-\varepsilon,S+\varepsilon]$ of $J$ for some $\varepsilon > 0$.

Then $S+\varepsilon \in I$, contradicting $\sup(I) = S$.

wh0
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  • This is pretty cool! Do you know other examples where such a method applies? – Gabriel Romon Mar 08 '14 at 21:06
  • I found a good answer here. I knew it from a "inductive" proof of Heine-Borel theorem and the question leads me to think of compactness of $[a,b]$. – wh0 Mar 09 '14 at 02:46
  • Ah thanks, I saw nice proofs of well-known theorems in the link! Nevertheless, I don't get how this is related to compactness... We only use here the Least Upper Bound Property. Therefore I can't think of any generalization to metric spaces other than $\mathbb R$. – Gabriel Romon Mar 09 '14 at 09:09
  • Yes. I read the link and also found that we don't need compactness – wh0 Mar 09 '14 at 09:12