What is the norm of the dual space $H^1(\Omega)'$? I am working on the Bidomain-Model which, during a time interval [0,T], describes the electrical behaviour of the myocardial muscle considered as $\Omega \subset \mathbb{R}^3$.
This model has partial differential equations which involve the intra- and extracellular voltages, $u$ and $v$, which are functions of this type: 
$$ u=u(t,x):[0,T]\times\Omega\rightarrow\mathbb{R}$$
$$ v=v(t,x):[0,T]\times\Omega\rightarrow\mathbb{R}$$ 
Now, as functions of time,
$$ u,v \in X=L^2\left(0,T,H^1(\Omega)\right)$$
with norm $\| u \|_{_{X}} = \left(\int_{_{[0,T]}}\| u(t) \|^{^2}_{_{H^1(\Omega)}} dt \right)^{1/2}$.
Their time derivatives
$$ \dot u,\dot v \in Y=L^2\left(0,T,H^1(\Omega)' \right)$$
with norm $\| u \|_{_{Y}} = \left(\int_{_{[0,T]}}\| u(t) \|^{^2}_{_{H^1(\Omega)'}} dt \right)^{1/2}$.
Further, for any fixed $t_0 \in [0,T]$, 
$$\| u(t_0) \|_{_{H^1(\Omega)}}^{^2} = \| \dot u(t_0,x) \|_{_{L^2(\Omega)}}^{^2} + \| u(t_0,x) \|_{_{L^2(\Omega)}}^{^2} = \langle u(t_0,x), u(t_0,x) \rangle_{_{H^1(\Omega)}}$$
but 

Which is $\| u(t_0,x) \|_{_{H^1(\Omega)'}}$?
  Is it $$\|u(t_0)\|_{_{H^1(\Omega)'}}^{^2} = \sup_{\xi \in H^1(\Omega)} \left\{ \langle  u(t_0,x), \xi(t_0,x) \rangle_{_{H^1(\Omega)}} \, | \, \| \xi(t_0,x)\|_{_{H^1(\Omega)}}^{^2} \le 1 \right\}$$
  as this post suggests?

 A: Yes, the dual-norm is that one!
Furthermore, you should consider following:
First: The embedding
$$ H^1(\Omega)\hookrightarrow L^2(\Omega). $$
This is because
$$ \| \cdot \|_{_{H^1(\Omega)}}^{^2} := \| D_{_t}^1\cdot \|_{_{L^2(\Omega)}}^{^2} +\| \cdot \|_{_{L^2(\Omega)}}^{^2} $$
is a "stronger" norm than $ \|\cdot\|_{_{L²(\Omega)}}$.
Second: it follows from the Fréchet-Riesz representation theorem that for the mentioned real-Hilbert-spaces:
$$ H^1(\Omega) \cong H^1(\Omega)^*$$
They are isometrically isomorphic by
$$ \Phi : H^1(\Omega) \rightarrow H^1(\Omega)^*$$
$$ \qquad \qquad u \quad \mapsto \,\langle \cdot, u\rangle_{_{H^1(\Omega)^*}}$$
Last but not least, regarding the spaces
$$ X:= L^2\left(0,T,H^1(\Omega)\right) \quad \text{and} \quad Y:= L^2\left(0,T,H^1(\Omega)^* \right)$$
we have a function
$$ \Psi : L^2 \left( 0,T,H^1(\Omega) \right) \rightarrow L^2 \left( 0,T,H^1(\Omega)^*\right)$$
$$ \qquad \qquad u \quad \mapsto \,\langle \cdot, u\rangle_{_{L^2\left(0,T,H^1(\Omega)^*\right)}}$$ 
which is an isometrical isomorphism and thus
$$ L^2 \left( 0,T,H^1(\Omega) \right) \cong L^2 \left( 0,T,H^1(\Omega)^*\right). $$ 
Good luck in your work!
