Proving that a subset is compact in $L^2(0,1)$ I have posted a related question earlier but like to make the question more specific now so that one may help point out any mistake I possibly make.
Consider the following two subsets in $L^2(0,1)$:
${\cal H}:= \{g \in L^2(0,1) \;|\; g:\text{non-decreasing}\}$
$ {\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 and are chosen such that the set ${\cal G}$ is nonempty. 
Obviously, ${\cal G} \subseteq {\cal H}$.
I claim that $cl({\cal G})$ is weakly compact and $cl({\cal G}) \subseteq {\cal H}$ where $cl()$ referes to the closure of the set.
My reasoning is based on the Banach-Alaoglu theorem, which states that 
"Let $X$ be the dual to some separable Banach space $Z$, $X = Z^*$. Then any bounded subset $M$ of $X$ is precompact in the weak-* topology, i.e. any sequence in $M$ has a weak-* convergent subsequence"
Given that any element in ${\cal G}$ is bounded by $\sqrt{c_2}$, we may apply the above theorem to conclude that ${\cal G}$ is weakly precompact, which by definition leads to $cl({\cal G})$ is weakly compact.
Since any sequence in ${\cal G}$ will always converge to a non-decreasing function, i.e. the limit points are in ${\cal H}$, we can conclude that the closure $cl({\cal G}) \subseteq {\cal H}$.
I just like to check if there is anything I might overlook in the above argument. Thanks for pointing out if any!
 A: It is ok but you should give some details on the fact that $\mathrm{cl}(G)\subset H$. That is, you should prove that the weak limit of a sequence of nondecreasing functions is nondecreasing. 
First of all, we deal here with equivalence classes up to almost everywhere (a.e.) equivalence, so the nondecreasing assumption is to be interpreted as holding up to an exceptional null set. This is conveniently characterized in integral form as follows: 
$$
f\text{ is "a.e. nondecreasing" }\ \Leftrightarrow \ \int f\left( -\frac{d\phi}{dx} \right)\, dx \ge 0 \quad \forall \phi\in C^\infty_c,\, \phi \ge 0.$$
Here $C^\infty_c$ denotes smooth functions with compact support in $(0,1)$ (so in particular, functions that vanish at $0$, $1$). 
You can now show that this "a.e. nondecreasing" condition is preserved by weak limits. By the way, the condition $\int f\, dx = c_1$ is also preserved by weak limits. The condition on the norm, on the other hand, is only partially preserved, as the norm of a weak limit can be strictly lower than the norm of the terms of the sequence. Summing up, you get that
$$
\mathrm{cl}(G)\subset \left\{ f\in L^2(0,1),\ f \text{ is a.e. nondecreasing},\ \int f\, dx=c_1,\ \int f^2\, dx \le c_2 \right\}.$$
