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 $H$).

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?

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 $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V)$), then is $$\lVert{u}\rVert_{C^0([0,T];V)} \leq C\lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;H)}?$$ (Notice the norm on $u'$ is taken with respect the weaker space $H$). The answer to this question may depend on the following question: is there a continuous embedding $$W(V,H) \hookrightarrow C([0,T];V)?$$

We know: $$W(V,V^*) \hookrightarrow C([0,T];H)$$ $$W(V,V) \hookrightarrow C([0,T];V)$$ but how about the intermediary case? My guess is there is no such embedding. But the first question may still be true.

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 $H$).

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. Is there a continuous embedding $$W(V,H) \hookrightarrow C([0,T];V)?$$

We know: $$W(V,V^*) \hookrightarrow C([0,T];H)$$ $$W(V,V) \hookrightarrow C([0,T];V)$$

but how about the intermediary case? My guess is there is no such embedding. In this case, can anyone answer this question: if $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V)$, then is $$\lVert{u}\rVert_{C^0([0,T];V)} \leq C\lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;H)}?$$ (Notice the norm on $u'$ is taken with respect the weaker space $H$). The answer to my first question obviously answers this.

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 $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V)$), then is $$\lVert{u}\rVert_{C^0([0,T];V)} \leq C\lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;H)}?$$ (Notice the norm on $u'$ is taken with respect the weaker space $H$). The answer to this question may depend on the following question: is there a continuous embedding $$W(V,H) \hookrightarrow C([0,T];V)?$$

We know: $$W(V,V^*) \hookrightarrow C([0,T];H)$$ $$W(V,V) \hookrightarrow C([0,T];V)$$ but how about the intermediary case? My guess is there is no such embedding. But the first question may still be true.