Consider the heat equation on the interval [0,1]
\begin{equation} u_t = u_{xx} \end{equation} \begin{equation} u(0) = u_0 \in H^2(0,1) \end{equation} with boundary conditions \begin{equation} u_{x}(0)=u_x(1)=0 \end{equation}
where $ H^2(0,1)$ is the space of twice weakly differentiable functions on $(0,1)$. If $(u_0)_x(0)=(u_0)(1)=0$ then $ \|u(\cdot,t)\|_{H^2(0,1)}$ is uniformly bounded for all $t\geq 0 $. However, does a uniform bound also hold if $u_0$ does not satisfy the Neumann boundary condition but $u_0 \in H^2(0,1)$?
It seems to me that such a uniform bound should hold since we have that $(\cdot)_{xx}$ with the Neumann BC generates a analytic semigroup, hence $\|u(\cdot,t)\|_{H^2(0,1)}$ is bounded for each $t > 0 $. Therefore, the only possibility is that $\|u(\cdot,t)\|_{H^2(0,1)}$ blows up near $t=0$. But we already know $\|u_0\|_{H^2(0,1)}$ is bounded. Hence, $ \|u(\cdot,t)\|_{H^2(0,1)}$ is uniformly bounded for all $t\geq 0 $. Is this argument is right? otherwise what can go wrong?