Let $1<p<\infty$, and $f\_n$ be a bounded sequence in $C(0,T;H^2(0,L))$. It looks obvious to me that $f\_n^p$ is also bounded in $C(0,T;H^2(0,L))$. When we take the derivative of $f^p(t)$ twice we get $$(f^p)\_{xx}(t)=p(p-1)f^{p-2}(t)(f\_x(t))^2+pf^{p-1}(t)f\_{xx}(t).$$
Since for every $t\in[0,T]$,$f(t)$ and $f\_x^2(t)$ are in $C(0,L)$, we have $(f^p)\_{xx}(t)$ in $H^2(0,L)$ using above formula. Does this prove the statement?



$f_n$ is bounded in $C(0,T;H^2(0,L))$ so is $f_n^p$?
Yes, your justification is correct but needs to be expanded upon to be rigorous. In particular, it is important that not only are $f$ and $f\_x$ continuous, but their $C^0$ norm is controlled by their Sobolev norm, so that a uniform bound can be derived. By Sobolev embedding theorem (since $k = 2 > n/2$ with $k$ the derivative order and $n = 1$ the dimension of the spatial domain), then there is a constant $C$ such that
$$
\|f(\cdot, t)\|\_{C^0(0,L)} \le C \| f(\cdot, t)\|\_{H^2(0,L)}
$$
for every $t$. Since $f(\cdot, t) \in C((0,T); H^2)$ then $\| f(\cdot, t)\|\_{H^2(0,L)} \le M$ for some $M$ uniformly in $t$. Therefore
$$
|(f^p)\_{xx}(x,t)| \le p (p-1) (CM)^{p-2} |f\_x(x,t)|^2 + p (CM)^{p-1} |f\_{xx}(x,t)|,
$$
and Sobolev embedding can again be applied instead to $f\_x$ to show that its $C^0(0,L)$ norm is bounded uniformly in $t$. Carrying out the inequalities and integrating gives you a universal bound on $\|(f^p)\_{xx}(\cdot, t)\|\_{L^2(0,L)}$ that is a polynomial expression in $C,M,p$. 


