# $f_1,f_2,\ldots$ continuous on $[0,1]$ s.t $f_1 \geq f_2 \geq \ldots$ and $\lim_{n\to\infty}f_n(x)=0$ .

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.

• Look at Dini's theorem on Wikipedia. – KCd Mar 18 '12 at 5:42
• 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. – Michael Hardy Mar 18 '12 at 19:02
• Just fixed a horrible typo in my answer. If you read it when it said "$\le$" and were confused, try again now. It says "$\ge$". – 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.
• @MichaelHardy : Sir , can you tell me why $C_{n,\epsilon }$ are compact ?? – Theorem Jun 9 '12 at 9:26
• 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 $[0,1]$. – Michael Hardy Jun 9 '12 at 17:18
• 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. – 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 $L=0$.