For the parabolic equation $$u_t + f(u)_x - u_{xx} = 0$$ one has $$\Vert u(t,\cdot) \Vert_{L^2(\mathbb R)} + 2\int_0^t \Vert u_x(s,\cdot) \Vert_{L^2} ds \le \Vert u(0,\cdot) \Vert_{L^2(\mathbb R)}.$$ If $t \le T$ (for a fixed $T>0$), does the estimate above imply also $$\Vert u_x (t,\cdot) \Vert_{L^2(\mathbb R)} \le C$$ (without time integration)?
In other words, one can also ask: if $\int_0^T f(s) ds \le C$ is it true that $f(t) \le C$ for all $t \in [0,T]$?
In the case of linear heat equation, we write $$ u(x,t) = \int_{\mathbb{R}} \Gamma(x-y,t) u_0(y) dy.$$ Here $\Gamma(x,t)$ is the standard heat kernel. Note that $$ \left|\left(\frac{\partial}{\partial x}\right)^k \Gamma(x,t) \right| \leq C_k t^{-1/2-k/2} \exp \left( -C |x|^2/t\right) $$ for some constant $C>0$.
Hence by Young's convolution inequality, we have $$ \Vert\nabla u(\cdot,t) \Vert_{L^2(\mathbb{R})}\leq \Vert{\nabla \Gamma(\cdot,t)}\Vert_{L^1(\mathbb{R})}\Vert{u_0}\Vert_{L^{2}(\mathbb{R})}. $$ A change of variable shows that $$ \Vert{\nabla \Gamma(\cdot,t)}\Vert_{L^1(\mathbb{R})} \leq C t^{-1/2}$$ for some constant $C>0$.
Hence we get $$ \Vert\nabla u(\cdot,t) \Vert_{L^2(\mathbb{R})}\leq C t^{-1/2}\Vert{u_0}\Vert_{L^{2}(\mathbb{R})} $$ for all $t>0$.
From this point of view, I think that we cannot expect uniform bound in time.
