In Spivak's chapter on uniform convergence he asks to prove the following

THEOREM Let $\{f_n\}$ be sequence of continuous functions that converge pointwise to $0$ over $[a,b]$. If $0\leq f_{n+1}\leq f_n$ for each $x$ and $n$, then convergence is actually uniform over $[a,b]$.

Now, he asks, as he did in other occasions, to argue by contradiction and use Bolzano Weierstrass to find an "appropriate sequence $\{x_n\}$". I'm guessing he wants me to find a sequence that goes to $0$ but $f_n(x_n)\not \to 0$. I honestly didn't look at that option, but I wrote the following direct proof :

PROOF (This was awfully wrong)

I also don't see why it is essential that the $f_n$ are continuous. Could you provide a proof using Bolzano Weierstrass?

With this, it seems one can prove Dini's theorem, which seems an immediate result:

THEOREM Let $f_n\to f$ pointwise and monotonically over $[a,b]$, with each $f_n$ continuous, and $f$ continuous. Then $f_n\to f$ uniformly.

PROOF Assume $\{f_n\}$ increasing, and set $g_n=f-f_n$. Then the $g_n$ are continuous, $g_n\to 0$ pointwise and $$0\leq g_{n+1}\leq g_n$$ By the above, $g_n\to 0$ uniformly over $[a,b]$, that is, $f_n\to f $ uniformly over $[a,b]$. If $\{f_n\}$ is decreasing, consider $\{-f_n\}$.

Then Spivak asks

$(1)$ What if $f$ is not continuous? $(2)$ What if we replace $[a,b]$ with $(a,b)$?

  • $\begingroup$ The problem in the picture is that $\mu$ may depend on $n$. $\endgroup$ – Davide Giraudo Nov 7 '12 at 20:58
  • $\begingroup$ In the first proof, I don't understand why do we have such a $N$; if we can provide such an integer, the proof would be finished. $\endgroup$ – Davide Giraudo Nov 7 '12 at 21:42
  • $\begingroup$ @DavideGiraudo Isn't that because the $f_n$ converge pointwise to $0$ over $[a,b]$? Then taking $\mu \in[a,b]$ such that $f_n(\mu)=\sup f_n$ it must be the case $\lim f_n(\mu)=0$. $\endgroup$ – Pedro Tamaroff Nov 7 '12 at 21:43
  • $\begingroup$ Where do you take the supremum? And this $\mu$ has no reason to be the same for different integers. $\endgroup$ – Davide Giraudo Nov 7 '12 at 21:44
  • $\begingroup$ @DavideGiraudo Nevermind. I see my mistake. I will try and write a proof using Bolzano Weiertrass. I had a feeling the proof couldn't be right. $\endgroup$ – Pedro Tamaroff Nov 7 '12 at 21:54

Assume that we can find $\delta>0$, a subsequence $\{f_{n_k}\}$ and a sequence $\{x_{n_k}\}$ such that $f_{n_k}(x_{n_k})\geqslant \delta$ for all $k$. Bolzano-Weierstrass theorem ($[a,b]$ is compact) allows us to extract of $\{x_{n_k}\}$ a subsequence, denoted $\{t_j\}$, converging to some $t$. We have for all integers $n$ and $m$, denoting $g_k$ the sequence indexed by the integers appearing in $t_k$, $$\delta\leqslant g_{m+n}(t_{m+n})\leqslant g_n(t_{m+n}).$$ Now we fix $n$, and take the $\limsup_{m\to +\infty}$. This gives for all integer $n$, $$\delta\leqslant\limsup_{k\to +\infty}g_k(t_k)\leqslant g_n(t),$$ a contradiction.

In the second theorem, take $a=0, b=1$.

  • If we don't assume $f$ continuous, consider $f_n(x):=x^n$.
  • The same counter-example considering $(0,1)$.
  • $\begingroup$ Thanks. Why is my first proof wrong? (The second one, I think is correct.) $\endgroup$ – Pedro Tamaroff Nov 7 '12 at 21:39
  • $\begingroup$ Yes, the second one is correct. $\endgroup$ – Davide Giraudo Nov 7 '12 at 21:45
  • $\begingroup$ Actually, we don't really need $\limsup$: from $\delta\leqslant g_n(n+m)$, take $\lim_{m\to\infty}$ and use continuity of $g_n$ to get $\delta\leqslant g_n(t)$ for all $n$. $\endgroup$ – Davide Giraudo Nov 8 '12 at 9:11

Here is a direct proof:

THEOREM Suppose $\{f_n\}$ is a sequence of continuous functions from $[a,b]$ to $\Bbb R$ that converge pointwise to a continuous function $f$ over $[a,b]$. If $f_{n+1}\leq f_n$, then convergence is uniform.

PROOF We set $g_n=f_n-f$ and note that $g_n\geq g_{n+1}$ and the $g_n$ are continuous, converging pointwise to $0$. For a given $\epsilon>0$. Consdier the (relatively) open sets (because of continuity of the $g_n$) $O_n=\{x\in [a,b] :g(x)<\epsilon\}$. Note that since $g_n\geq g_{n+1}$ we have $O_n\subset O_{n+1}$. Given $x\in[a,b]$ there is an $n$ such that $g_n(x)<\epsilon$; whence $\bigcup_{n\in\Bbb N}O_n=[a,b]$. But since $[a,b]$ is compact there exists a finite set $K=\{1,\dots,m\}$ such that $\bigcup_{k=1}^m O_{n_k}=[a,b]$. But since $O_n\subset O_{n+1}$ the greatest element of $K$, call it $\ell$, is such that $O_\ell =[a,b]$. And we're done: for every $n\geq \ell$ we have $g_n(x)<\epsilon$; as desired. $\blacktriangle$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.