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Suppose $S(r)$ is set parametrised by $r \in [0,T]$. Let $\phi_t^s : H^1(S(t)) \to H^1(S(s))$ is a linear homeomorphism.

Suppose $\lVert \cdot \rVert_{H^1(S(t))}$ and $\lVert \phi_t^s(\cdot) \rVert_{H^1(S(s))}$ are equivalent norms on $H^1(S(t))$. So their dual norms are also equivalent.

The norm of $f \in H^{-1}(S(s))$ can be given as $$\sup_{w \in H^1(S(s))} \frac{\langle f, w \rangle_{H^{-1}(S(s)), H^1(S(s))}}{\lVert w\rVert_{H^1(S(s))}} = \sup_{v \in H^1(S(t))} \frac{\langle (\phi_t^s)^*f, v \rangle_{H^{-1}(S(t)), H^1(S(t))}}{\lVert \phi_t^sv\rVert_{H^1(S(s))}}$$ where the star denotes the adjoint.

I can derive this by just using the fact that \phi_t^s is a homeomorphism and the adjoint identity. Do I need any equivalence of dual norms at all?

Suppose that $$\lVert \phi_t^sv\rVert_{H^1(S(s))} \leq C\lVert v \rVert_{H^1(S(t))}$$

This, and the equivalence of $H^1$ norms implies $$\lVert (\phi_t^s)^*\rVert_{\mathcal{L}(H^{-1}(S(s)), H^{-1}(S(t))} \leq C$$

I still don't need any equivalence of norms to get this. Can somebody show me how to do it correctly?

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  • $\begingroup$ $\phi_t^s$ being a linear homeomorphism implies that $\|-\|$ and $\|\phi_t^s(-)\|$ are equivalent ... $\endgroup$ – martini Oct 25 '12 at 16:44
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We don't need the assumption that the norm are equivalent or that $\lVert\phi_t^s v\rVert_{H^1(S(s))}\leq C\lVert v\rVert_{H^1(S(t))}$, as it's already implied by the property of linearity and homeomorphism. Indeed, if $L\colon E\to F$ is linear and continuous, where $E$ and $F$ are normed spaces, then we can find $C>0$ such that $$\lVert Lx\rVert_F\leq C\lVert x\rVert,\quad x\in E.$$

To see the equality between the two suprema, we just need the fact that $\phi_t^s$ is a bijective correspondance.

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  • $\begingroup$ You say linearity + homeomorphism implies norm equivalence. Is the converse true? Does norm equivalence (in the form given above) imply homeomorphism? $\endgroup$ – soup Oct 29 '12 at 12:01
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    $\begingroup$ If $C^{-1}\lVert x\rVert\leq \lVert Lx\rVert\leq C\lVert x\rVert$ and $L$ is linear, then $L$ is continuous and injective. But not necessarily bijective (take $E=\ell^2$ and $L$ the shift to the right). $\endgroup$ – Davide Giraudo Oct 29 '12 at 12:05

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