2
$\begingroup$

Let's consider the following function space $$ K = \{ f \in C[0, 1] | \int_{0}^{t}{|f(s)| dt} \leq t^{4}, \forall t \in [0,1] \}$$

I would like to establish, whether this space is compact or not. Since we'are working at $C[0,1]$ it can be done more or less straighforward, for example, applying Arzela-Ascoli theorem, which states that:

$K \subset C[0,1]$ is a compact subspace iff $K$ is closed, bounded and equicontinuous, i.e. $\forall \epsilon >0, \exists \delta>0, \forall x,y$ so that $|x-y| < \delta$, $|f(x)-f(y)| < \epsilon$ $\forall f \in K$

Well, despite, i got some troubles with straightforward check. Are there any hints the might help? Or maybe, is there a different approach to the given problem?

Any help would be much appreciated.

$\endgroup$
  • $\begingroup$ what is you distance over $C[0,1]$ ? $\endgroup$ – Sebastien Nov 29 '15 at 12:56
  • $\begingroup$ You're speaking about Arzela-Ascoli so I guess that it's the uniform distance. $\endgroup$ – Sebastien Nov 29 '15 at 12:57
  • $\begingroup$ @Sebastien Well, it matters much, for instance, let's consider $C[0, 1]$ equipped with a uniform norm, i. e. $|| f ||_{\infty} = \sup_{x \in [0, 1]} {|f(x)|}$ $\endgroup$ – hyperkahler Nov 29 '15 at 12:58
2
$\begingroup$

The function $ 4t^3$ satisfies $$\int_0^t |4t^3| ds = t^4$$ for all $t$. Thus $$L = \{ f\in C[0,1] : |f(t)| \le 4t^3 \}$$ is a subset of $K$. While in this $L$ there is a canonical family which is not equicontinuous, which is $$\{ 4t^3, 4t^4, 4t^5, \cdots\}.$$ Thus $K$ is also non-compact.

$\endgroup$
  • $\begingroup$ Well, how did you manage to derive that $L$ is as big as the underlying set? Anyway, it seems to be that for any polynomial bound the situation would not change. $\endgroup$ – hyperkahler Nov 29 '15 at 13:49
  • 1
    $\begingroup$ @Arteom : The term big here should be interpreted vaguely. That could be make precise by writing down some isometric embedding to $K$, but we don't need that. Note that it is easy to write down a sequence of functions in $L$ which converge pointwisely to a discontinuous function. $\endgroup$ – user99914 Nov 29 '15 at 13:54
  • $\begingroup$ Well, i wonder, whether $L$ is a subset of $K$. For example. let $t=\frac{2}{3}$, then $\int_{0}^{\frac{2}{3}}{f(s)ds} \leq {(\frac{2}{3})}^{4}$, which clearly doesn't hold for any $||f||_{\infty} \leq 1$ on $x \in [\frac{1}{2}; \frac{2}{3}]$. $\endgroup$ – hyperkahler Dec 1 '15 at 19:42
  • 1
    $\begingroup$ @Arteom : I think so, I've changed the answer to a more direct one. $\endgroup$ – user99914 Dec 2 '15 at 3:06
  • $\begingroup$ Well, also, it's possible to take the sequence $nt^{n-1}$, $n \geq 4$ and it will also work, since it's not bounded in uniform norm. $\endgroup$ – hyperkahler Dec 9 '15 at 21:20

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.