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Let $(M, \mu, d)$ be a geodesically complete non-compact Riemannian manifold such that measure $\mu$ is volume doubling, i.e. \begin{equation}\label{VD}\mu(B(x, 2r))\leq C\mu(B(x, r))\end{equation} for some constant $C>0$ and also $M$ satisfying the $L^{2}$-Poincare inequality $$\frac{1}{\mu(B(x, r))}\int_{B(x, r)}{|u-u_{B(x, r)}|^{2}d\mu}\leq cr^{2}\frac{1}{\mu(B(x, \delta r))}\int_{B(x, \delta r)}{|\nabla u|^{2}d\mu}$$ for all $u\in W^{1, 2}(B(x, r))$ with $\delta>1$ and $\nabla u$ being the weak gradient of $u$.

For a fixed point $x_{0}\in M$ and $\alpha, \beta\in \mathbb{R}_{+}$ consider the annuli $$P_{\alpha, \beta}=\{x\in M:\alpha<d(x, x_{0})<\beta\}.$$

Question: Is there a statement known, such that under these assumptions (or even stronger), any function $u\in L_{loc}^{2}(P_{\alpha, \beta})$ with $$\int_{P_{\alpha, \beta}}{|\nabla u|^{2}d\mu}<\infty$$ can be extended to a function $\widetilde{u}\in L_{loc}^{2}(M)$ such that $$(*)\int_{M}{|\nabla \widetilde{u}|^{2}d\mu}\leq C\int_{P_{\alpha, \beta}}{|\nabla u|^{2}d\mu}?$$

The closest statement I found is from "On extensions of Sobolev functions defined on regular subsets of metric measure spaces" by P. Shvartsman. In this paper he proofs that considering a regular set $S$, i.e. a set such that there are constants $\theta_{S}\geq 1$ and $\delta_{S}>0$ such that for every $x\in S$ and $0<r\leq\delta_{S}$ $$\mu(B(x, r))\leq \theta_{S}\mu(B(x, r)\cap S),$$ then any function $u\in L^{2}(S)$ such that $u_{1, S}^{\#}\in L^{2}(S)$ where $$u_{1, S}^{\#}(x):=\sup_{r>0}\frac{r^{-1}}{\mu(B(x, r))}\int_{B(x, r)\cap S}{|u-u_{B(x, r)\cap S}|d\mu}$$ can be extended to a function $\widetilde{u}\in CW^{1, 2}(M)$ such that $$\|\widetilde{u}\|_{CW^{1, 2}(M)}\leq C(\|u\|_{L^{2}(S)}+\|u_{1, S}^{\#}\|_{L^{2}(S)}),$$ where $CW^{1, 2}(M)$ is the Calderon-Sobolev space which coincides with the classical Sobolev space $W^{1, 2}(M)$ if one assumes volume doubling and $M$ satisfying the $L^{2}$-Poincare inequality.

Thanks in advance for your help!

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    $\begingroup$ If the constant $C$ is allowed to depend on the manifold $M$ and the annulus $P_{\alpha,\beta}$, then, since $\overline{P}_{\alpha,\beta}$ is compact, it suffices to do this locally. Therefore, an extension theorem for a domain in $\mathbb{R}^n$ suffices. Stein's book, Singular Integrals and Differentiability Properties of Functions, contains such a theorem, $\endgroup$ – Deane Yang Aug 6 at 17:18
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If the annulus is small, then it is basically an Euclidean annulus and there is an extension operator for Sobolev spaces. However, if the annulus is large is may happen that it goes around a "neck" in a manifold and meets along the boundary as the picture shows.

enter image description here

Then a smooth function on the annulus can approach to $0$ from one side of the common part of the boundary and to $1$ from the other side. Such a function has no Sobolev extension at all.

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  • $\begingroup$ Thank you very much for your answer! Is there a condition on $M$ or the annuli known, such that there is an extension and the inequality $(*)$ holds? $\endgroup$ – Shaq155 Aug 9 at 10:44

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