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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?

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  • $\begingroup$ Hint: $H$ stands for Hilbert $\endgroup$ – user251257 Apr 18 '16 at 10:54
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    $\begingroup$ Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) $\endgroup$ – Martin Sleziak Apr 18 '16 at 15:54
  • $\begingroup$ Thank you @MartinSleziak! I edited my question, hope this helps me get a satisfying answer (: $\endgroup$ – scjorge Apr 18 '16 at 21:30
  • $\begingroup$ By definition norm of dual of the normed space it is that you say, but it is not very useful as well write. It is useful to characterize the $H^1$-norm through the Fourier transform, so also see who is the norm of the dual. As in this thread math.stackexchange.com/questions/1694713/… $\endgroup$ – user288972 Apr 20 '16 at 10:43
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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!

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