# Compactness of the $K \subset C[0,1]$

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.

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

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.
• 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. – hyperkahler Nov 29 '15 at 13:49
• @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. – user99914 Nov 29 '15 at 13:54
• 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 $x \in [\frac{1}{2}; \frac{2}{3}]$. – hyperkahler Dec 1 '15 at 19:42
• 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. – hyperkahler Dec 9 '15 at 21:20