3
$\begingroup$

My question comes from the necessity of proving that one of the hypothesis in the Arzela Ascoli theorem fails in the example:

$f_n(x)=\frac{1}{1+(x-n)^2}; x\in [0,\infty)$

I have tried proving it is not uniformly equicontinuous nor uniformly bounded but keep failing, so my only assumption is that Arzela Ascoli cannot be applied because $[0,\infty)$ is closed but not bounded, so it is not compact in $\mathbb{R}$. Yet, I cannot think of a way to show the theorem fails in this example.

$\endgroup$
  • $\begingroup$ Note that we don't know that the theorem necessarily fails (or rather, that the conclusion of the theorem is false for $f_n$). However, that is indeed the case, as seen by José Carlos Santos' answer. What goes wrong, is that the function is just moving along the x-axis instead of getting closer to the zero-function (the limit). This cannot happen on a compact set. $\endgroup$ – Milten Oct 24 '19 at 7:40
  • $\begingroup$ Not really answering your question, but from an intuitive point of view it's kind of clear that a theorem proving compactness of something that extends X requires compactness of X itself. Indeed, the continuous functions on X extend X in exactly the sense required for this intuitive argument, by the Urysohn lemma. $\endgroup$ – Bananach Oct 24 '19 at 7:56
  • $\begingroup$ (well, there are still the uniform continuity and boundedness assumptions that I'm glossing over here, but after all you already have a precise and correct answer) $\endgroup$ – Bananach Oct 24 '19 at 7:59
3
$\begingroup$

This sequence is equi-continuous and uniformly bounded. The only reason Arzela-Ascoli Theorem is not applicable here is because $[0,\infty)$ is not compact.

Note that $0 \leq f_n(x) \leq 1$. So the sequence is uniformly bounded. Let su prove equi-continuity: $|f_n(x)-f_n(y)| \leq \frac {|x-y| (|(x-n)+(y-n)|} {(1+(x-n)^{2}) (1+(y-n)^{2})}$. Use the inequality $|a| \leq \frac 1 2 (1+a^{2})$ with $a=x-n$ and $a=y-n$ to see that the sequence is equi-continuous.

Note that $f_n(x) \to 0$ for each $n$ whereas $f_n(n)=1$. This shows that the sequence has no uniformly convergent subsequence.

$\endgroup$
  • $\begingroup$ Perfect! Just one more thing, why can't we extract a convergent subsequence from $f_n$? What I mean is: How can I explicitly prove that without compactness, we cannot extract a convergent subsequence from a uniformly bounded and uniformly equicontinuous $f_n$? $\endgroup$ – PLanderos33 Oct 24 '19 at 7:39
  • 1
    $\begingroup$ @PLanderos33 Since $f_n \to 0$ point-wise if there is a convergent subsequence the limit has to be $0$. But if $|f_{n_k} (x)-0| <1$ for all $k \geq k_0$ and for all $x$ we get a contradiction by putting $x=n_k$. Hence no subsquence converge uniformly. $\endgroup$ – Kavi Rama Murthy Oct 24 '19 at 7:43
  • $\begingroup$ Lastly, I am having trouble applying the inequality that you mentioned to show it is equicontinuous, any extra hints? $\endgroup$ – PLanderos33 Oct 24 '19 at 8:20
  • $\begingroup$ From the inequality I mentioned you get $|f_n(x)-f_n(y)| \leq \frac 1 4 |x-y|$ for all $n$. [Because $\frac {|x-n| {1+(x-n)^{2}} \leq \frac 1 2$ and $\frac {|y-n| {1+(y-n)^{2}} \leq \frac 1 2$. From this we get equi-continuity. @PLanderos33 $\endgroup$ – Kavi Rama Murthy Oct 24 '19 at 8:24
  • 1
    $\begingroup$ @KaboMurphy Don't you mean $f_n(x)\rightarrow 0$ as $n\rightarrow \infty$ for each $x$ whereas $f_n(n)=1$ showing that $\lim_{n\rightarrow\infty}\sup_{x\in[0\infty)}|f_n(x)-0|=1\neq 0$? $\endgroup$ – Peter Melech Oct 24 '19 at 8:28
4
$\begingroup$

You have a sequence $(f_n)_{n\in\mathbb N}$ of functions which is uniformly bounded and equicontinuous. Asserting that the theorem fails in that situation means that the conclusion of the theorem doesn't hold (and not that one of the hypothesis doesn't hold); in other words, $(f_n)_{n\in\mathbb N}$ has no subsequence which is uniformly convergent.

In order to see why, not that, for each $x\in[0,\infty)$, $\lim_{n\to\infty}f_n(x)=0$, but $(\forall n\in\mathbb N):f_n(n)=1$.

$\endgroup$

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.