How to fix a proof of Dini's Lemma

Question

I am aware of 2 proofs of Dini. One is by contradiction and Bolzano-Weierstrass, and one is by an open covers definition. I decided to try to make a 'direct' proof without using 'every open cover has a finite subcover', but I struggle to finish it. For reference:

(Dini) Suppose $F_n,F:[a,b]→ℝ$ are continuous functions such that for every $x∈[a,b]$, $F_n(x)\downarrow F(x)$ as $n→∞$. Prove that in fact, $F_n → F$ uniformly.

Note first that without loss, $F=0$ and $F_n\geq 0$. I want to argue that I can control $|F_n(x) - F_n(y)| ≤ ω(|x-y|)$ with a modulus of continuity $ω$ independent of $n$; using this I can then create a grid of points $(a_i)$ $δ/2$ apart so that $ω(δ)<ε$, and choose an $M$ so that for $m>M$, $\sup_i F_m(a_{i})<ε $, then conclude via $$ F_m(x) ≤ F_m(a_{i,x}) + |F_m(a_{i,x}) - F_m(x)| ≤ \sup_i F_m(a_{i}) +ω(δ) < 2ε $$ where $a_{i,x}$ is some $a_i$ close to $x$.

I unfortunately don't see how to prove that $ω$ can be chosen independently of $n$; is this possible?

Answer 1

[Continued: ... indulge me.]

There is agreement now that this question is really about Dini's theorem and compactness arguments. Mr. Khor wanted to try for a proof based on using a single modulus of continuity for the whole sequence, not to avoid compactness arguments--just for a different perspective.

I think all proofs of Dini's theorem look the same when you stare long enough, so his proposed proof might shift that perspective around, but it won't be very radical.

The big four compactness arguments on the real line [HB, BW, Cousin-coverings (CC), nested-intervals (NI)] are the major tools one learns in a first course. BW generalizes to compact metric spaces and HB to compact topological spaces, but CC and NI don't survive past $\mathbb{R}^n$.

There are two more "compactness arguments" on the real line, although I would describe them more as "back-to-basics." For completeness I will sketch them here. These arguments are quite nice when they can be used, but don't expect all problems to offer a tidy way of using them. Keep HB, BW, etc. handy.

Theorem (Dini) Suppose $F_n: a,b]\to\mathbb{R}$ are continuous functions decreasing pointwise to zero. Then the convergence is uniform.

Proof using monotone-convergence property: Let $\epsilon>0$. We claim we can find an integer $M$ so that $|F_M(x)|<\epsilon$ for all $a\leq x \leq b$ and that would prove the theorem.

Let $x_0=a$, select the first integer $N_0$ so that $F_{N_0}(x_0)< \epsilon$. If $N_0$ doesn't work for our $M$ then there is an $x_1>x_0$ so that $F_{N_0}(x )<\epsilon$ for all $x_0\leq x<x_1$ and $F_{N_0}(x_1)=\epsilon$. (This by continuity.) Now choose $N_1$ the same way, as the first integer with $F_{N_0}(x_0)< \epsilon$. Etc.

Do this inductively and within a finite number of steps we reach an integer $N_k$ that we can take as our $M$. If not then we have two increasing sequences $\{x_k\} $ and $\{N_k\}$, the first bounded and the second has $N_k\to \infty$.

By the monotone-convergence property $\lim_{k\to \infty}x_k=z$ exists and $a<z\leq b$. By the same arguments we have seen repeatedly, there is a first $N$ so that $F_N(z)<\epsilon$ and a contradiction is easily deduced.

Proof using least upper bound property: Let $\epsilon>0$. We claim we can find an integer $M$ so that $|F_M(x)|<\epsilon$ for all $a\leq x \leq b$ and that would prove the theorem.

Define $S$ as the set of all points $s\in [a,b]$ such that there is an integer $N$ (which may depend on $s$) so that for every $a\leq t \leq s$ the inequality $F_N(t)<\epsilon$ must hold. Argue (i) $S$ is nonempty and bounded. (ii) Define $z=\sup S$. (iii) Prove that $z\in S$. (iv) Prove that $z<b$ is impossible.

[We are using the least upper bound property to obtain (ii). I prefer to call it the last point argument, since we are imagining there is a last point $z$ beyond which we cannot prove the statement that we want to prove.]

The same details that worked before are used to prove (i), (iii), and (iv) [as expected of course].

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A more general point of view on Dini's theorem.

For fans of Dini's theorem there is another way of looking at the proof that, I think, clarfies what is going on. Use this definition and theorem.

Definition Let $F$, $F_n$ be real-valued functions defined on $\mathbb{R}$. We say $F_n\to F$ uniformly locally at a point $x$ if, for every $\epsilon>0$ there is an integer $N$ and $\delta>0$ so that $$ |F_n(t)-F(t)|< \epsilon $$ for all $n\geq N$ and all $|t-x|<\delta$.

Theorem Let $F$, $F_n$ be real-valued functions defined on $\mathbb{R}$. Suppose that $F_n\to F$ uniformly locally at every point $x$ in a compact set $K$. Then $F_n\to F$ uniformly on $K$.

Now what is Dini's theorem? Under the assumptions of Dini's theorem all functions are continuous on a compact interval $[a,b]$ and the convergence is monotone and pointwise to $F$. Just show, quickly, that in that case $F_n\to F$ uniformly locally at every point $x$ in $[a,b]$ and apply the theorem. While Dini's theorem seems to be about continuity and monotonic convergence it just happens to be a local uniform condition. All compactness arguments are about going from the local to the global and that is all that is happening here.

Answer 2

Don't have any ideas on your proposed proof, other than that already suggested by Mr. Sinclair, namely that whatever argument you use will certainly invoke compactness.

What I can do is add to your collection of proofs. You know a Bolzano-Weierstrass argument and Heine-Borel argument for Dini's theorem. In Exercise 9.4.7 Elementary Real Analysis I proposed as a problem constructing proofs based on the other compactness arguments. I don't recall what I had in mind but maybe it is time to do it myself.

I think of Heine-Borel as the tool for serious scholars who dislike inelegance and distrust anything that can not be generalized to extremes. Bolzano-Weierstrass is more get-your-hands-dirty and should appeal to engineers and plumbers. Cousin's covering lemma is, because of its Belgian origin, more effete and esoteric, with suggestions of Hercule Poirot, absinthe, Balkan Sobranies in long cigarette holders, etc. The nested interval argument is just down on the farm, basic, get-the-work done attitude. So, for your collection, here are the other proofs.

Theorem (Dini) Suppose $F_n:[a,b] \to\mathbb{R}$ are continuous functions decreasing pointwise to zero. Then the convergence is uniform.

Proof using Cousin coverings: Let $\epsilon>0$. Define $\mathcal{C}$ as the collection of all subintervals $[u,v]$ of $[a,b]$ with the property that, for some integer $n$ ($n=n(u,v)$) it is true that $F_n(x)< \epsilon$ for all $u\leq x \leq v$. This collection $\mathcal{C}$ is a full cover in the Cousin sense, since if $x\in [a,b]$ there is an integer $n$ so that $F_n(x)<\epsilon$ and so, by continuity a positive $\delta$ with $F_n(t)<\epsilon$ for all $|t-x|<\delta$. Thus all closed subintervals of $(x-\delta,x+\delta)$ belong to $\mathcal{C}$ .

By Cousin's lemma there is a partition $\{[u_i,v_i];i=1,2,\dots,n\}$ of $[a,b]$ contained in $\mathcal{C}$ . Take $N=\max\{n(u_i,v_i):i=1,2,\dots, n\}$ and any point $x$. Select the interval $[u_i,v_i]$ to which $x$ belongs and check that $$ F_m(x)\leq F_{n(u_i,v_i)}(x)<\epsilon$$ for all $m\geq N$.

Proof using a nested sequence of intervals: If the convergence is not uniform then there is a $\epsilon_0>0$ so that the following statement is true:

For infinitely many $n$ the inequality $f_n(x) < \epsilon _0$ fails at some point $x\in [a,b]$ which may depend on $n$.

Split the interval into two equal halves and choose $[a_1,b_1]$ as the half for which this is true:

For infinitely many $n$ the inequality $f_n(x) < \epsilon _0$ fails at some point $x\in [a_1,b_1]$ which may depend on $n$.

A standard nested sequence of intervals is defined by induction and this shrinks to a single point $z$ in $[a,b]$. We know that $F_N(z)< \epsilon_0$ for some $N$ and so, by continuity there is a positive $\delta$ with $F_N(t)<\epsilon_0$ for all $|t-x|<\delta$. For large enough $M$ the interval $[a_M,b_M] \subset(z-\delta,z+\delta)$ and one has, for all $t$ in the interval $[a_M,b_M]$, that $$ F_n(t)\leq F_N(t) < \epsilon_0$$ for all $n$ greater than both $M$ and $N$. This contradicts the choice of interval $[a_M,b_M]$. The contradiction completes the proof.