$f_n$ is bounded in $C(0,T;H^2(0,L))$ so is $f_n^p$?

Question

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?

Answer 1

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 $C,M,p$.