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For a bounded connected domain with the interior cone property, and $1\le p<n$, the mean-zero form of the Poincaré inequality isNo zero boundary trace is imposed. Subtracting the mean is essential because nonzero constants have zero gradient.

Suppose the estimate failed. Subtract means and normalize to obtain $v_j\in W^{1,p}(\Omega )$ withThis sequence is bounded in the Sobolev space $W^{1,p}$. The allowed Rellich-Kondrachov compactness theorem on cone domains provides a subsequence converging strongly in $L^p(\Omega )$ to $v$. Its mean is zero and its norm is one. For any compactly supported smooth test function $\varphi$,Thus every weak derivative of $v$ vanishes. To show this implies constancy, mollify on any ball compactly contained in $\Omega$. The resulting smooth function has zero gradient and is constant on the smaller ball. Passing to its $L^p$ limit makes $v$ constant almost everywhere on that ball. Overlapping balls have the same constant, and connectedness connects any two points by a chain of such balls. Hence $v$ is constant almost everywhere throughout $\Omega$. Its zero mean makes it zero, contradicting norm one. This proves the stated Poincaré inequality. The frequently used stronger Sobolev-Poincare formulation also follows: with $p^{\ast }=np/(n-p)$, the cone-domain Sobolev embedding theorem gives $\Vert w{\Vert }_{p^{\ast }}\le C_S(\Vert w{\Vert }_p+\Vert \nabla w{\Vert }_p)$. Insert $w=v-v_{\Omega }$ and the estimate just proved to obtainThus both the same-exponent and critical-exponent mean-zero versions are covered. Connectedness cannot be dropped without subtracting a separate mean on every component. The usual meaning of the stated range is $1\le p<n$; $W^{1,p}$ with $p<1$ is not part of this Sobolev-space formulation.