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Consider a bounded and connected open set (with regular boundary) $\Omega\subset\mathbb{R}^d$ and a family of $\mathscr{C}^1$-diffeomorphisms $(\varphi_t)_{t\in[0,\alpha]}$ all in $\mathscr{C}^0(\overline{\Omega})$. I would like to know some sufficient/necessary assumptions on the previous family to insure that $(C_{\Omega_t})_{t\in[0,\alpha]}$ remains bounded, where $C_{\Omega_t}$ is the Poincaré-Wirtinger constant for $\Omega_t:=\varphi_t(\Omega)$ that is the best positive number such as for all $u\in H^1(\Omega_t)$ \begin{align*} \left\|u-\frac{1}{|\Omega_t|}\int_{\Omega_t} u\right\|_{L^2(\Omega_t)} \leq C_{\Omega_t} \|\nabla u\|_{L^2(\Omega_t)}. \end{align*} A sufficient condition seems to be that $(\varphi_t)_{t\in[0,\alpha]}$ is bounded in $W^{1,\infty}(\Omega)$ and that all the considered diffeomorphisms preserve the volume since in this case, by a simple change of variable we get that $C_{\Omega_t} \leq C_{\Omega} \displaystyle\sup_{t\in[0,\alpha]}\|\nabla \varphi\|_{\infty}$.

Another approach would be to argue that (assuming $(\varphi_t)_{t\in[0,\alpha]}$ is bounded in $L^\infty$) all the sets $\Omega_t$ are included in some big open set $O$ and try to use some extension argument to bound $(C_{\Omega_t})_{t\in[0,\alpha]}$ in terms of $C_O$. I tried for example the extension operator $P_t:H^1(\Omega_t) \rightarrow H^1(O)$, since I knew that the constant of continuity of $P_t$ may be computed using a partition of unity that would be hopefully transported by diffeomorphism, but since $\|\nabla P_t(u)\|_2$ is not controlled by $\|\nabla u\|_2$ (but only by $\|u\|_2+\|\nabla u\|_2$), this approach won't work.

Would be glad to hear any advice / comment / idea !

I added the tag " spectral theory ", since the Poincaré-Wirtinger constant is also the inverse of the smallest positive eigenvalue of the Neumann-Laplacian on $L^2(\Omega_t) / \mathbb{R}$. I hope this is will attract some comments !

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Have you seen ? – Willie Wong Oct 17 '12 at 14:47
Well, what can I say ? Thank you ! In fact, in my case I figure out a proof that works, but it's good to know this general result. – xounamoun Oct 18 '12 at 15:10

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