Define $W(X,Y) = \{ u \in L^2(0,T;X) \mid u' \in L^2(0,T;Y)\}$
where $u'$ is the usual weak derivative.


Let $V \subset H \subset V^\*$ be a Hilbert triple. If $u \in W(V,V)$ (so $$u \in L^2(0,T;V) \quad\text{with}\quad u' \in L^2(0,T;V)$$), then is $$\lVert{u}\rVert\_{C^0([0,T];V)} \leq C\left(\lVert{u}\rVert\_{L^2(0,T;V)} + \lVert{u'}\rVert\_{L^2(0,T;H)}\right)?$$ (Notice the norm on $u'$ is taken with respect the weaker space $H$). 


We know that 
$$W(V,V) \hookrightarrow C([0,T];V).$$
So the quantities in the desired inequality make sense, but does the inequality hold?



Bounding $\lVert{u}\rVert_{C^0([0,T];V)} \leq C\left(\lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;H)}\right)$?
As mentioned by @TaQ, the embedding $W(V,H) \hookrightarrow C([0,T];V)$ is, in general, not true. However, if your estimate would be true, you can extend the embedding operator from the dense subspace $H^1(0,T;V)$ to $L^2(0,T;V) \cap H^1(0,T; H) = W(V,H)$. This yields a contradiction.


