1
$\begingroup$

I am attempting to solve the following problem:

Let the sequence of continuous functions $\{x_{n}(t) \}_{n=1}^{\infty}$, $0 \leq t < \infty$ be uniformly bounded on $t \in [0, \infty)$ and on each bounded interval $[0,A]$ also satisfy the Lipschitz condition: $\forall t_{1}, t_{2} \in [0,A]$, $|x_{n}(t_{1})-x_{n}(t_{2})|\leq L_{A} |t_{1}-t_{2}|$, where the Lipschitz constant $L_{A}$ is independent of the index $n \geq 1$ but depends on $A$.

(a) Prove that the sequence $x_{n}$ is pre-compact (meaning its closure is compact) in the space of bounded continuous functions $C_{B}[0, \infty)$ with the supremum norm, and thus has a limit point $x \in C_{B}[0, \infty)$. (OP Note: I wonder if, in this case, the supremum norm should have been used in the definition of the Lipschitz condition above?)

(b) Find an example showing that $\{x_{n}\}_{n=1}^{\infty}$ may not be compact in $L^{1}(0, \infty)$.

For Part (a), this is what I've done so far:

Since $x_{n}(t)$ is uniformly bounded on each bounded interval, it satisfies the Lipschitz condition $\exists L_{A}$ s.t. $\Vert x(t^{\prime})-x(t) \Vert_{\infty} \leq L_{A} \Vert t^{\prime} - t \Vert_{\infty}$, then letting $\delta = \frac{\epsilon}{L_{A}}$ responds to the $\epsilon > 0$ challenge regarding the equicontinuity at each point in $[0,A]$, since $L_{A}$ is independent of $n$ as is $\epsilon$ and $\delta$.

By the Arzela-Ascoli Theorem, $\exists $ a subsequence of $\{x_{n}\}$ that converges pointwise on all of $[0,A]$ to a real-valued function $x(t)$ on $[0,A]$.

To extract the subsequence we need, first consider the case where $A=1$. Then, since $\{x_{n}(t)\}_{n}^{\infty}$, $0 < t < \infty$ is uniformly bounded on $t \in [0, \infty)$, and Lipschitz on $[0,1]$, it satisfies the conditions of Arzela-Ascoli. So, $\exists$ a convergent subsequence $\{x_{1_{n}}\}$.

Then, on $[0,2]$, $\{x_{1_{n}} \}$ is still convergent, so it is bounded. (OP Note: I'm not sure if this is the right justification, and I think around here, my argument gets a bit sloppy, so I need some help tightening this part up.) Therefore, $\exists $ a convergent subsequence $\{x_{2_{n}} \}$ by the Bolzano-Weierstrass Theorem.

Then, on $[0,3]$, $\{x_{2_{n}} \}$ is still convergent, so it is bounded. Therefore, $\exists$ a convergent subsequence $\{x_{3_{n}} \}$ by Bolzano-Weirstrass.

Continuing on in this fashion, we obtain the convergent subsequences $\{x_{1_{n}}\}$, $\{x_{2_{n}}\}$, $\cdots$, $\{x_{j_{n}}\}$, $\cdots$.

By construction, $\{x_{i_{n}}\}$ is a subsequence of all $\{x_{k_{n}}\}$ where $i < k$.

Since the $\{ x_{k_{n}}\}$ are bounded, by Bolzano-Weirstrass, $\{x_{i_{n}}\}$ converges.

Finally, $\{x_{n_{n}}\}$ is a subsequence of all the subsequences, so $\lim_{n \to \infty}x_{n_{n}}$ exists on $[0, \infty) = \lim_{A \to \infty}[0,A]$.

Is this enough in order to show that the sequence is pre-compact? I.e., to show that its closure is compact? Or do I need any additional details? If so, what are they?

I know this problem requires a diagonalization argument, but I don't know if I have it down quite the way it's supposed to be, so I would appreciate any help in fixing this up. Also, I don't really see how the supremum norm came into play here, which makes me feel like I did something very, very wrong.

As for part (b), well...I know the closed unit ball is not compact in $L^{1}$, but that's not the limit of a sequence in $C_{B}$...

Otherwise, I'm completely lost on part (b), and would appreciate any help you could give me.

Thank you.

$\endgroup$
  • 1
    $\begingroup$ For part b), there is a trivial answer : there is no reason for the $x_n$ to be in $L^1$ ! (take $x_n=1$) $\endgroup$ – Tryss Apr 18 '16 at 22:13
  • $\begingroup$ @Tryss, can you help at all with the other part? $\endgroup$ – ALannister Apr 18 '16 at 23:19
  • $\begingroup$ Houston, we have a problem. If we endow $C_B[0,\infty)$ with the supremum norm, such a sequence need not be precompact. By Ascoli, you have a subsequence that converges locally uniformly on $[0,\infty)$, but it need not have a subsequence converging uniformly on all of $[0,\infty)$ (let for example $f(t) = e^{-t^2}$ for $t\in \mathbb{R}$ and $x_n(t) = f(t-n)$ for $t\in [0,\infty)$; the sequence satisfies the premises - the Lipschitz constant is even independent of $A$ - and converges locally uniformly to $0$, but it has no uniformly convergent subsequence). $\endgroup$ – Daniel Fischer Apr 20 '16 at 11:01
  • $\begingroup$ @DanielFischer, it's possible that's what my prof meant - converging locally uniformly. $\endgroup$ – ALannister Apr 20 '16 at 11:10
  • $\begingroup$ @DanielFischer, what about the other question? The one I had tagged you on? $\endgroup$ – ALannister Apr 20 '16 at 11:10
1
$\begingroup$

I think what you have written is sufficient. You have shown that the closure of the sequence in question has some limit point, which means that the closure of the sequence is sequentially compact, which is the same as compactness. As for where you used the $\sup$ norm, it was used in the Arzela-Ascoli theorem, wasn't it? After all, the convergences are happening in the supremum norm, and that is the convergence given by the Arzela Ascoli theorem as well.

For part b), there is no need for the sequence to be in $L^1$ : If the sequence is not in $(0,\infty)$, then as Tryss has said above, you have trivial examples like the one he gave, having no limit point in $L^1$ by virtue of not even being in $L^1$. The set he has given has a limit point in the set of bounded and continuous functions on $(0,\infty)$. It's correct, but it's not illustrative.

We can make up a more illustrative example, like this: take some $L^1$ function of your choice, I'll take the function $f=\frac{1}{\pi}e^{-\frac{x^2}{2}}$, which has integral $1$ on the interval $(0,\infty)$. Now take $x_n=nf$, just an integer multiple of the function. The integral of $x_n$ is increasing, hence you see that it is not having a limit point in $L^1$, but the point is the sequence itself is in $L^1$. That is the example of a sequence in $L^1$ that does not have a limit point in $L^1$, or in the set of bounded and continuous function for that matter, showing that both these sets are not compact in themselves i.e. precompact.

$\endgroup$

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.