# Best constant for Poincaré inequality on spheres

I am interested in the following Poincaré-type inequality, $$\int_{S(r)} \lvert u-\bar{u}\rvert^2 d\sigma \leq C(N) \int_{S(r)} |u_{\theta}|^2 d\sigma$$ where $$\bar{u} = \frac{1}{\lvert S(r)\rvert}\int_{S(r)} u d\sigma$$ and $$u_{\theta}$$ denotes the tangential derivative of $$u$$. The domain $$S(r)$$ is just the $$N$$-dimensional sphere with radius $$r$$. The function $$u$$ can be assumed to be smooth for the purpose of this question.

In the case $$S(r)$$ is $$1$$ dimensional (i.e. a circle), then the constant $$C(1)=1$$ and this is the Wirtinger's inequality. Are there any references where I can find the best constant in higher dimensions?

The best constant is just the multiplicative inverse of the smallest positive eigenvalue of the Laplacian on the sphere. On $$\mathbb{S}^N$$ this the smallest eigenvalue is $$N$$, so $$C(N)$$ in that case equals $$1/N$$.
You can figure out the appropriate $$r$$ scaling yourself.
• Incidentally, this fact is generally true. If you have a closed connected Riemannian manifold, the global Poincare inequality like you stated has the best constant equal to the inverse of smallest positive eigenvalue of the Laplace-Beltrami operator (with sign condition so the spectrum is non-negative). If you have a compact connected manifold with boundary, the principal eigenvalue of the Dirichlet Laplacian provides the best constant for the Poincare inequality $\|u\|_{L^2}^2 \lesssim \|\nabla u\|_{H^1}^2$ for functions vanishing on the boundary. Commented Mar 9, 2022 at 21:15
• The determination of the eigenvalue for the sphere usually doesn't do the Rayleigh quotient. One proves first that the spherical harmonics (as restrictions of harmonic polynomials on $\mathbb{R}^{N+1}$) are a complete system of orthogonal functions in $L^2$. That the polynomials are harmonic and homogeneous implies that the spherical harmonics are also eigenfunctions of the spherical Laplacian, with explicit eigenvalues. By the spectral theorem these spherical harmonics must be all the eigenvalues, and by inspection you see the smallest one. Commented Mar 10, 2022 at 15:24