# Gårding's inequality

In mathematics, Gårding's inequality is a result that gives a lower bound for the bilinear form induced by a real linear elliptic partial differential operator. The inequality is named after Lars Gårding.

## Statement of the inequality

Let $\Omega$ be a bounded, open domain in $n$-dimensional Euclidean space and let $H^{k}(\Omega )$ denote the Sobolev space of $k$-times weakly differentiable functions $u\colon \Omega \rightarrow \mathbb {R}$ with weak derivatives in $L^{2}(\Omega )$. Assume that $\Omega$ satisfies the $k$-extension property, i.e., that there exists a bounded linear operator $E\colon H^{k}(\Omega )\rightarrow H^{k}(\mathbb {R} ^{n})$ such that $Eu\vert _{\Omega }=u$ for all $u\in H^{k}(\Omega )$.
Let L be a linear partial differential operator of even order 2k, written in divergence form
    $(Lu)(x)=\sum _{0\leq |\alpha |,|\beta |\leq k}(-1)^{|\alpha |}\mathrm {D} ^{\alpha }\left(A_{\alpha \beta }(x)\mathrm {D} ^{\beta }u(x)\right),$
and suppose that L is uniformly elliptic, i.e., there exists a constant θ > 0 such that
    $\sum _{|\alpha |,|\beta |=k}\xi ^{\alpha }A_{\alpha \beta }(x)\xi ^{\beta }>\theta |\xi |^{2k}{\mbox{ for all }}x\in \Omega ,\xi \in \mathbb {R} ^{n}\setminus \{0\}.$
Finally, suppose that the coefficients A$_{αβ}$ are bounded, continuous functions on the closure of Ω for |α| = |β| = k and that
    $A_{\alpha \beta }\in L^{\infty }(\Omega ){\mbox{ for all }}|\alpha |,|\beta |\leq k.$
Then Gårding's inequality holds: there exist constants C > 0 and G ≥ 0
    $B[u,u]+G\|u\|_{L^{2}(\Omega )}^{2}\geq C\|u\|_{H^{k}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{k}(\Omega ),$
where
    $B[v,u]=\sum _{0\leq |\alpha |,|\beta |\leq k}\int _{\Omega }A_{\alpha \beta }(x)\mathrm {D} ^{\alpha }u(x)\mathrm {D} ^{\beta }v(x)\,\mathrm {d} x$
is the bilinear form associated to the operator L.

## Application: the Laplace operator and the Poisson problem

Be careful, in this application, Garding's Inequality seems useless here as the final result is a direct consequence of Poincaré's Inequality, or Friedrich Inequality. (See talk on the article).
As a simple example, consider the Laplace operator Δ. More specifically, suppose that one wishes to solve, for f ∈ L$^2$(Ω) the Poisson equation
    ${\begin{cases}-\Delta u(x)=f(x),&x\in \Omega ;\\u(x)=0,&x\in \partial \Omega ;\end{cases}}$
where Ω is a bounded Lipschitz domain in R$^n$. The corresponding weak form of the problem is to find u in the Sobolev space H$_0$$^1$(Ω) such that
    $B[u,v]=\langle f,v\rangle {\mbox{ for all }}v\in H_{0}^{1}(\Omega ),$
where
    $B[u,v]=\int _{\Omega }\nabla u(x)\cdot \nabla v(x)\,\mathrm {d} x,$
    $\langle f,v\rangle =\int _{\Omega }f(x)v(x)\,\mathrm {d} x.$
The Lax–Milgram lemma ensures that if the bilinear form B is both continuous and elliptic with respect to the norm on H$_0$$^1$(Ω), then, for each f ∈ L$^2$(Ω), a unique solution u must exist in H$_0$$^1$(Ω). The hypotheses of Gårding's inequality are easy to verify for the Laplace operator Δ, so there exist constants C and G ≥ 0
    $B[u,u]\geq C\|u\|_{H^{1}(\Omega )}^{2}-G\|u\|_{L^{2}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{1}(\Omega ).$
Applying the Poincaré inequality allows the two terms on the right-hand side to be combined, yielding a new constant K > 0 with
    $B[u,u]\geq K\|u\|_{H^{1}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{1}(\Omega ),$
which is precisely the statement that B is elliptic. The continuity of B is even easier to see: simply apply the Cauchy–Schwarz inequality and the fact that the Sobolev norm is controlled by the L$^2$ norm of the gradient.