Could anybody please help me check whether the following statement is true?
For $T>0$ and $\Omega = (0,1)$, let $f, g \in C^1([0,T];L^2(\Omega))\bigcap C([0,T];H^1(\Omega))$ such that $$\min_{t \in [0,T]}\min_{x \in [0,1]} g(t,x) \geqslant c > 0$$ and $$\sup_{t \in [0,T]} \left[\|f(\cdot,t)\|_{H^1}+\|g(\cdot,t)\|_{H^1}+\|f'(\cdot,t)\|_{L^2}+\|g'(\cdot,t)\|_{L^2}\right] \leqslant R<\infty$$ then $\dfrac{f|f|}{g} \in C^1([0,T];L^2(\Omega)).$
My attempt was the following. The continuity in time follows from the fact that every function involved is continuous and the bottom does not vanish. Membership in $L^2(\Omega)$ follows by the embedding $H^1(\Omega) \in L^\infty(\Omega) \subset L^4(\Omega)$. Indeed, for each $t$, $$\int_0^1 \left|\dfrac{f|f|}{g}\right|^2dx = \int_0^1 \dfrac{f^4}{g^2}dx \leqslant \dfrac{1}{c^2}\|f\|_{L^4}^4 < \infty.$$ For the time derivative, we compute: $$\dfrac{\partial}{\partial t} \dfrac{f|f|}{g} = \dfrac{2|f|f_tg - f|f|g_t}{g^2}$$ which is again continuous due to continuity of all the functions involved. For membership in $L^2(\Omega)$ we use again the same embedding and compute $$\int_0^1 \dfrac{4f^2f_t^2}{g^2}dx \leqslant \dfrac{4\|f\|_{L^\infty}^2}{c^2}\|f_t\|_{L^2}^2 < \infty,$$ and $$\int_0^1 \dfrac{f^4g_t^2}{g^4}dx \leqslant \dfrac{\|f\|_{L^\infty}^4}{c^4}\|g_t\|_{L^2}^2 < \infty,$$ which means that both functions in the time derivative belong to $L^2$, hence their sum is also in $L^2.$
Is there any mistake in this proof? And if not, is there any way to conclude it without having to perform so many computations?