Getting the bound $\frac{1}{h}\int_0^{T-h}\int_t^{t+h}\int_\Omega |\nabla u(\tau)| |\nabla u(t+h) - \nabla u(t)|\;dxd\tau dt \leq C$

Question

Let $u \in L^2(0,T;H^1(\Omega)) \cap L^\infty(0,T;L^2(\Omega)).$

Is it possible to find the following bound: $$\frac{1}{h}\int_0^{T-h}\int_t^{t+h}\int_\Omega |\nabla u(\tau,x)| |\nabla u(t+h,x) - \nabla u(t,x)|\;dxd\tau dt \leq C\quad \forall h \in [0,T]$$ where $C$ is allowed to depend on (appropriate norms of) $u$ but should not depend on $h$.

Answer 1

Since there is no information about how $\nabla u$ changes with respect to $t$, we can't hope to obtain any cancellation in $|\nabla u(t+h,x) - \nabla u(t,x)|$. Just estimate it by $|\nabla u(t+h,x)| + |\nabla u(t,x)|$.

Using the notation $U(t) = \|\nabla u(t,\cdot)\|_{L^2}$ and the Cauchy-Schwarz inequality, we estimate the given integral by
$$\frac{1}{h} \int_0^{T-h} \int_t^{t+h} U(\tau) (U(t+h)+U(t))\,d\tau\,dt \tag{1}$$ Since $U\in L^2(0,T)$, its one-sided Hardy-Littlewood maximal function $$ M_U(x) = \sup_{h>0} \frac{1}{h} \int_t^{t+h} U(\tau) \,d\tau $$ is also in $L^2(0,T)$ (with an appropriate norm bound). Thus, (1) is estimated by $$ \int_0^{T-h} M_U(t) (U(t+h)+U(t)) \,dt$$ which is integral of the product of two $L^2$ functions, hence finite.