Gårding's inequality

Remarks

The form $B$ is the sesquilinear form of the elliptic operator $Lu = \sum (-1)^{\abs{\alpha}} \partial^\alpha(A_{\alpha\beta}\partial^\beta u)$ in divergence form. For constant coefficients and only top-order terms, the inequality follows from Plancherel's theorem: the Fourier transform turns $\operatorname{Re} B(u, u)$ into $\int \operatorname{Re}\sigma(\xi)\abs{\hat u(\xi)}^2 \, d\xi \ge \theta \int \abs{\xi}^{2k}\abs{\hat u(\xi)}^2 \, d\xi$. For variable coefficients, one freezes the coefficients on small pieces of $\Omega$, using uniform continuity, and glues with a partition of unity. Lower-order terms are absorbed with the interpolation inequality $\norm{u}_{H^j} \le \varepsilon\norm{u}_{H^k} + C_\varepsilon\norm{u}_{L^2}$ for $j < k$.

Example: for $k = 1$ and $B(u, u) = \int_\Omega \abs{\nabla u}^2 \, dx$, the form of $-\Delta$, one has $B(u, u) = \norm{u}_{H^1}^2 - \norm{u}_{L^2}^2$, so $c = 1$ and $G = 1$ work.

Ellipticity cannot be dropped. In $\R^2$, the form $\int_\Omega \abs{\partial_1 u}^2 \, dx$ is not elliptic, and functions that oscillate fast in the $x_2$ direction make $\norm{u}_{H^1}$ large while the form and $\norm{u}_{L^2}$ stay bounded.

Going deeper

For a student

Adding $G$ to the operator, the form $B(u, v) + G\ip{u}{v}$ becomes coercive on $H^k_0(\Omega)$. The Lax–Milgram theorem then gives a unique weak solution of $(L + \lambda)u = f$ for every $f$ and every $\lambda \ge G$. Combined with the compact embedding of $H^k_0(\Omega)$ in $L^2(\Omega)$, this yields the Fredholm alternative and a discrete spectrum for the Dirichlet problem. The constant $G$ cannot in general be taken to be $0$, because $L$ may have eigenvalues with negative real part.