I am looking at a set of functions defined as follows
$ {\cal G}:=\{ g \in L^2(0,1) \;|\; g:\text{non-decreasing}, \int_0^1 g(\alpha)d\alpha = c_1, \int_0^1 g^2(\alpha)d\alpha=c_2\}$, where $c_1\in \Re$ and $c_2>0$ are some fixed values.
Assuming that $c_1$ and $c_2$ are chosen such that the above set is not empty, and $c_2>0$, I am trying to prove whether the set is compact or not.
I have been trying different arguments, but they are all quite sketchy. I wonder if anyone can give me some ideas of how to prove or disprove this rigorously.
thank you!
Such a set cannot be compact. Here is an example when $c_1=0, c_2 = 1.$ Then $\mathcal G$ contains all functions of form $$f_p(x)=\begin{cases} -\sqrt{\frac p{1-p}}& x\le p\\\sqrt{\frac{1-p}p}& x>p\end{cases},$$ for values of $p\in(0,1)$. If $\mathcal G$ were compact, there would be a subsequence of $p_n\to0$ of the sequence $1/n$ such that the corresponding $f_n$ converged to some $L^2$ limit $f$. What could $f$ limit be? It has the property that $\langle f,g\rangle=0$ for all $g$ with support in $[0,a]$ for any $a<1$, as you can (and should) verify by evalutating $\lim \langle f_n,g\rangle$ for such $g$. Since the set of such $g$ is dense in $L^2$ (I leave the why to you) we conclude that $f=0$. But then $\|f-f_n\|=1$ for all $n$, contrary to $f_n\to f$.
The same trick works for other $c_1,c_2$ values.