# Constant in Poincare Inequality

Let $I$-identity operator, $\Pi_N$ is the orthogonal projection in $L_2$ onto subspace by the first $N$ eigenfunctions of the Stokes operator in $\Omega$, $\alpha_j$ denotes the increasing sequence of the eigenvalues for the Stokes operator, $c>0$ is the some constant not depending on $N$. We know that $\Pi_N v = 0$.

How to derive the following inequality using the Poincare inequality? Which form of inequality is used here? $|v|_{L_2}=|(I-\Pi_N)v|_{L_2}\le c\alpha_N^{-1/2}|v|_{H^1}$.

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Can someone edit the tex, please? –  András Bátkai Mar 4 '11 at 15:28

This is a fairly standard stuff. Suppose that the Stokes operator $A=-\Delta$ is defined on smooth divergence-free vector fields $u$ which satisfy the standard no-slip boundary condition $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.u\right|_{\partial\Omega}=0.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1)$$ Let $H$ be the closure in $L^2$-norm of the space of smooth divergence-free vector fields with compact support: $$H=\overline{\{u=(u_1,\dots,u_d)\in (C_0^\infty(\Omega))^d:\ \mbox{div}\ u=0\}}^{L^2}$$
It is well known that $A$ gives rise to a self-adjoint operator with compact inverse on $H$ provided that $\Omega\subset \mathbb R^d$ is a bounded domain with Lipschitz boundary (see, e.g., Chapter 1 of Navier-Stokes Equations by Temam). Moreover, one can show that $D(A^{1/2})\subset (H^1(\Omega))^d$ and that, for any $u\in D(A^{1/2})$, $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\|A^{1/2}u\|_{L^2}^2=\|\nabla u\|_{L^2}^2.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(2)$$
Now, let $\{e_k\}_{k\in\mathbb N}$ be the orthonormal basis in $H$ which consists of eigenvectors of $A$. The required estimate is implied by (2) and the trivial inequality $$\|(I-\Pi_N)v\|_{L^2}^2=\sum\limits_{k\geq N} |(v,e_k)|^2\leq \frac{1}{\lambda_N}\sum\limits_{k\geq N}\lambda_k |(v,e_k)|^2=\frac{1}{\lambda_N}\|A^{1/2}v\|^2_{L^2}\qquad\qquad$$ which holds true for any $v\in D(A^{1/2})$ such that $\Pi_N v =0$.