Compactness of a subspace of $L^{2}([0,1])$

Question

Let's consider the following space $K \subset L^{2}[0, 1]$, consisting of fucntions $x(t)$ so that $\sin{t} \leq x(t) \leq t$. How to check, whether this subspace is compact in $L^{2}[0, 1]$ or not.

I suppose that it's not, since it's relatively 'big' to be a compact space. Since, the reasonable approach is to suggest a counterexample, including sequence of functions so that it's impossible to extract convergent subsequence.

Some useful techniques, such as Arzela-Ascoli theorem, do not help much, since it's impossible to make accurate estimates from the given data.

Are there any hints that might help?

Answer 1

Hint: First, let $E$ be the set of $f\in L^2([0,1])$ such that $|f|\le1$. Then $E$ is not compact (hint: trig functions...)

Now, there exists a rectangle $[a,b]\times[c,d]$ contained in $\{(t,s):0\le t\le 1,\sin(t)\le s\le t\}$. Hence, given a sequence $f_n$ in $E$ with no convergent subsequence you can construct a sequence in $K$ with no convergent subsequence, by simply...

Answer 2

Let $\{e_n :n\in N\}$ be an orthonormal basis for $L^2[1/2,1]$ with $n\neq m\to e_n\ne e_m,$ such that $A=\{\sup \{|e_n(x)| :x\in [1/2,1]\}<\infty$ for every $n$. (For example, a commonly used base of certain cos and sin functions.) $$\text { For }x\in [0,1/2) \text { let } f_n(x)=x.$$ $$\text {For } x\in [1/2,1] \; \text {let } f_n(x)=\frac {1}{2}(x+\sin x) +\frac {1}{2}\frac {e_n(x)}{A}\ ( 1/2- \sin (1/2)).$$ We have $f_n\in E$ because $x-\sin x$ is increasing. But in $ L^2[0,1]$ we have $\|f_n-f_m\|=\frac{1/2-\sin (1/2)}{A\sqrt 2}$ whenever $n\ne m.$