I am trying to prove the Remark (Remarque) 1.1 in a paper by P.A. Raviart published in 1970 about parabolic equations with a nonlinearity in the time derivative. The remark, where no computations are provided, recites:
Let $\Omega$ be a bounded domain in $\mathbb{R}^2$, let $T>0$ be given, let $1< \alpha <2$, and let $u \in L^\infty(0,T;L^\alpha(\Omega))$ be such that: $$ \dfrac{d}{d t}\left(|u|^\frac{\alpha-2}{2}u\right) \in L^2(0,T;L^2(\Omega)). $$
Then, we have that $\left(|u|^\frac{\alpha-2}{2} u\right) \in \mathcal{C}^0([0,T];L^2(\Omega))$ and that $u\in\mathcal{C}^0([0,T];L^\alpha(\Omega))$.
My attempt
(i) $\left(|u|^\frac{\alpha-2}{2} u\right) \in \mathcal{C}^0([0,T];L^2(\Omega))$.
To see this, observe that: $$ \int_{0}^{T} \int_{\Omega} \left||u|^\frac{\alpha-2}{2}u\right|^2 d x d t= \int_{0}^{T} \int_{\Omega} |u|^\alpha d x d t. $$
Since it was assumed that $u \in L^\infty(0,T;L^\alpha(\Omega))$, then the latter integral is finite. Hence, combining the latter with the assumption on the distributional derivative in time gives $$ \left(|u|^\frac{\alpha-2}{2}u\right) \in H^1(0,T;L^2(\Omega)) \hookrightarrow \mathcal{C}^0([0,T];L^2(\Omega)). $$
This completes the proof of (i).
(ii) An auxiliary result. Let $\alpha>1$. Then $$ |x-y|^\alpha \le \left||x|^\alpha - |y|^\alpha\right|, \quad\mbox{ for all } x,y \ge 0. $$
To show the auxiliary result assume, without loss of generality, that $x \ge y > 0$, and let $$ t:=\dfrac{x}{y} \ge 1. $$
Therefore, the sought inequality is equivalent to proving that $$ (t-1)^\alpha \le t^\alpha -1,\quad\mbox{ for all }t \ge 1. $$
Consider the function $f:[1,\infty) \to \mathbb{R}$ defined by $$ f(t):=(t-1)^\alpha-t^\alpha+1. $$
Observe that $$ f'(t)=\alpha (t-1)^{\alpha-1}-\alpha t^{\alpha-1}=\alpha((t-1)^{\alpha-1}-t^{\alpha-1}). $$
Since $\alpha-1>0$ and since $0\le t-1 <t$, the monotonicity of the power operator gives that $(t-1)^{\alpha-1} \le t^{\alpha-1}$, so that $f'(t)\le 0$ for all $t>1$. Since $f(1)=0$, we infer that $f(t)\le 0$ for all $t \ge 1$ and the proof of (ii) is complete.
(iii) $|u|\in\mathcal{C}^0([0,T];L^\alpha(\Omega))$.
We have to show that for each $t_0 \in [0,T]$, $$ \lim_{t\to t_0} \int_{\Omega} ||u(t)|-|u(t_0)||^\alpha d x =0. $$
By part (ii), we have that the Cauchy-Schwarz inequality and the triangle inequality give \begin{align*} &\int_{\Omega} ||u(t)|-|u(t_0)||^\alpha d x \le \int_{\Omega} \left||u(t)|^\alpha-|u(t_0)|^\alpha\right| d x\\ &=\int_{\Omega}\left|\left|\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|^{2/\alpha}\right|^\alpha-\left|\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|^{2/\alpha}\right|^\alpha\right| d x =\int_{\Omega}\left|\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|^2-\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|^2\right| d x\\ &=\int_{\Omega} \left|\left(\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|-\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|\right) \cdot \left(\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|+\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|\right)\right| d x\\ &\le \left\|\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|-\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|\right\|_{L^2(\Omega)} \left\|\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|+\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|\right\|_{L^2(\Omega)}\\ &\le \left\||u(t)|^\frac{\alpha-2}{2}u(t)-|u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right\|_{L^2(\Omega)} \left\|\left||u(t)|^\frac{\alpha-2}{2}u(t)\right|+\left||u(t_0)|^\frac{\alpha-2}{2}u(t_0)\right|\right\|_{L^2(\Omega)}. \end{align*}
Observe that the continuity in $[0,T]$ of $\left(|u|^\frac{\alpha-2}{2}u\right)$ implies the uniform boundedness of the second factor, as well as that the first factor tends to zero as $t \to t_0$. This completes the proof of (iii).
Now, I don't know how to go on. I observe that the conclusion is true if the target space $L^\alpha(\Omega)$ is replaced by $\mathbb{R}$. I cannot see how to extend the argument to the vector-valued case.
Could you please help me?