Let $f_1,f_2,\ldots$ be continuous functions on $[0,1]$ satisfying $f_1 \geq f_2 \geq \cdots$ and such that $\lim_{n\to\infty} f_n(x)=0$ for each $x$. Must the sequence $\left\{f_n\right\}$ converge to $0$ uniformly on $[0,1]$?

I want to say yes and I want to try to invoke Arzela-Ascoli Theorem. My intuition tells me that for large $n$ we have that the slope of $f_n$ on small enough interval has slope less than 1. So $f_n$ must be Lipschitz and hence equicontinuous. So it must have a uniformly convergent subsequence, but that gives us that $f_n$ must converge uniformly to $0$.

Any help would be appreciated.

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    $\begingroup$ Look at Dini's theorem on Wikipedia. $\endgroup$ – KCd Mar 18 '12 at 5:42
  • $\begingroup$ You can get by with a lot less than Arzela-Ascoli. There's a fairly simple way to do it, unless I've missed something in my answer below. $\endgroup$ – Michael Hardy Mar 18 '12 at 19:02
  • $\begingroup$ Just fixed a horrible typo in my answer. If you read it when it said "$\le$" and were confused, try again now. It says "$\ge$". $\endgroup$ – Michael Hardy Mar 18 '12 at 19:07

Let $C_{n,\varepsilon} = \{x\in[0,1] : f_n(x) \ge \varepsilon\}$. Then $C_{1,\varepsilon} \supseteq C_{2,\varepsilon}\supseteq\cdots$ and these are compact sets. Since $\lim\limits_n f_n=0$, their intersection is empty (assume $\varepsilon>0$, of course). Therefore the intersection of some finite subcollection is empty. The sequence of $C$s reaches $\varnothing$ after only finitely many steps. So the sequence of $f$s is uniformly $<\varepsilon$ after finitely many steps. Hence you have uniform convergence.

  • $\begingroup$ @MichaelHardy : Sir , can you tell me why $C_{n,\epsilon } $ are compact ?? $\endgroup$ – Theorem Jun 9 '12 at 9:26
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    $\begingroup$ The set $[\varepsilon,\infty)$ is closed. The inverse-image of a closed set under a continuous function is closed. A closed bounded set in the line (or in any Euclidean space) is compact. The set is bounded since it's a subset of $[0,1]$. $\endgroup$ – Michael Hardy Jun 9 '12 at 17:18
  • $\begingroup$ @MichaelHardy I think this explain clearly but i have a question. Why do we need to have the compactness restriction? can't it just closed or just bounded? As it seems the proof doesn't involved the property of compactness but if it is not compact the theorem is not true in general. Can you briefing explain the role of the compact set in the proof? $\endgroup$ – Mathematics Dec 8 '12 at 17:40
  • $\begingroup$ In Euclidean spaces such as the real line, closed and bounded is the same as compact. In some other metric spaces, a set can be closed and bounded but not compact, and in some such cases there is a family satisfying the finite intersection property but having empty intersection. For example, in the space $(0,1)$ with the usual metric, all of the sets $(0,x]$ are closed and bounded. And the intersection $\bigcap\limits_{x\in S}(0,x]$ is non-empty whenever $S$ is finite. But the intersection $\bigcap\limits_{x\in(0,1)}(0,x]$ is empty. $\endgroup$ – Michael Hardy Dec 8 '12 at 19:47

Define the sequence $a_n=\max\limits_{x \in [0,1]} f_n(x)$. The fact that $f_n$ is decreasing proves that $a_n$ is also decreasing and bounded by zero, thus convergent. Denote by $L=\lim a_n$. If $L=0$ then $f_n$ uniformly converges to zero.

For every $n$ there is an $x_n$ such that $f_n(x_n)=a_n$. Passing eventually to a subsequence we can say that $x_n$ is convergent and denote $x=\lim x_n$.

Pick $\varepsilon >0$. For $p$ large enough, from continuity of $f_n$ we get that $$ |f_n(x_{n+p})-f_n(x)|<\varepsilon $$ Therefore, we can write $$ f_{n+p}(x_{n+p})-f_n(x)\leq f_n(x_{n+p})-f_n(x) <\varepsilon$$ which is $$ a_{n+p}-f_n(x)<\varepsilon$$ Take $p \to \infty$ and get $$ L -f_n(x)<\varepsilon$$ Take $n \to \infty$ and get $L<\varepsilon$. Finally for $\varepsilon \to 0$ we get $L=0$.


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