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 $$(f^p)_{xx}(t)=p(p-1)f(t)(f_x(t))^2+pf^2(t)f_{xx}(t).$$ 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?

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?