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Hardy's inequality bounds averages or weighted functions by their original size or derivatives. For $p>1$ and $f\ge 0$, its integral form is ${\int }_0^{\infty }(x^{-1}{\int }_0^{x}f(t)dt{)}^{p}dx\le (p/(p-1){)}^p{\int }_0^{\infty }f(x{)}^{p}dx$. Spatial weighted forms include the Hardy inequality in Euclidean space.

The positive energy theorem concerns the total gravitational energy of isolated systems, rather than a pointwise gravitational energy density. Its standard asymptotically flat form isHere are the hypotheses used in a spinorial proof. Take smooth, complete, three-dimensional initial data in general relativity $(\Sigma ,h,k)$ with a spin structure, no inner boundary, and finitely many asymptotically Euclidean ends. Assume the usual differentiable falloff $h_{ij}-{\delta }_{ij}=O(r^{-1})$, $\partial h=O(r^{-2})$, $k_{ij}=O(r^{-2})$, with sufficiently decaying further derivatives and integrable matter densities. The Hamiltonian constraint and momentum constraint must hold, and the matter must satisfy the dominant energy condition. These hypotheses exclude a naked singularity or an unaccounted boundary flux from the argument. An oriented three-manifold admits a spin structure; in higher dimensions the spinorial proof needs this as an additional assumption. Work in signature $(-+++)$ and units $G=c=1$.

Let $n$ be the future unit normal, choose $k_{ij}=-g({\nabla }_{i}n,e_j)$, and define $\mu =T_{ab}{n}^a{n}^b$ and $J_i=-T_{ab}{n}^a{e}_i^b$. With this convention the Hamiltonian constraint and momentum constraint readThe dominant energy condition implies $\mu \ge |J{|}_h$. The asymptotic charges are the ADM energy and ADM momentum,These are boundary charges; a sign condition on $\mu$ alone does not visibly determine their sign.

Choose gamma matrices obeying $\{{\gamma }^a,{\gamma }^b\}=2g^{ab}$, with spatial matrices Hermitian and ${\gamma }^{\hat{0}}$ anti-Hermitian. The relevant derivative is the Sen spinor connection, not just the intrinsic spin connection:The symmetric extrinsic curvature makes $\mathcal{D}={\gamma }^i{D}_i^{(3)}-\frac{1}{2}(\mathrm{tr}k){\gamma }^{\hat{0}}$. This is an elliptic differential operator with the ordinary spatial Dirac operator as principal part. Solve the Witten spinor equation $\mathcal{D}ϵ=0$, with $ϵ\to {ϵ}_{\infty }$ constant at the end under consideration and zero asymptotic constants at any other ends.

The central algebraic ingredient is the Witten-Nester identity. Define $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$, the future-directed Dirac current $K^a=\overline{ϵ}{\gamma }^aϵ$, and the Nester two-formIn an orthonormal frame $K^0=|ϵ{|}^2$. Since $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$ is Hermitian with eigenvalues $\pm 1$ for any unit spatial vector $q$, $|\mathbf{K}|\le K^0$. Thus the Dirac current is future causal. Differentiate the Nester two-form by the product rule. The antisymmetric second-derivative terms reduce using the spinor curvature identityThe Clifford algebra and the first Bianchi identity turn the curvature contraction into the Einstein tensor. The spatial derivative terms give the difference of two squares. Projecting onto $\Sigma$ therefore givesThe constraints are exactly the normal components of the Einstein field equations needed here. Moreover,For a solution of the Witten spinor equation, both terms left in this divergence are nonnegative.

There is an essential analytic step: an asymptotically constant spinor field solving the Witten spinor equation must actually exist. Extend the chosen constant smoothly from infinity to a field ${ϵ}_0$. Then $\mathcal{D}{ϵ}_0=O(r^{-2})$ is square-integrable. For compactly supported fields $\eta$, the same Witten-Nester identity has zero boundary flux and givesThe asymptotically Euclidean Hardy inequality in Euclidean space, local elliptic estimates and this identity supply the coercive estimate in the weighted first-order Sobolev space of decaying corrections. One way to see why a compact-region error cannot spoil coercivity is to assume the estimate fails and take a normalized sequence whose Dirac operator tends to zero. Local compactness and the estimate on the ends produce a decaying homogeneous solution. Integrating the identity with cutoffs shows that such a solution is parallel for the Sen spinor connection. A parallel field tending to zero at infinity is zero, contradicting its normalization. The end estimates exclude escape of the normalized sequence to infinity.

More constructively, on this completed correction space solvefor every compactly supported $\chi$, using the Lax-Milgram theorem. Set $ϵ={ϵ}_0+\eta$. Since $\mathcal{D}$ is formally skew-adjoint in the positive spatial spinor inner product, the residual $\xi =\mathcal{D}ϵ$ is a square-integrable weak solution of $\mathcal{D}\xi =0$. Elliptic regularity makes it smooth. Cutoff integration of the identity excludes a nonzero square-integrable homogeneous solution: it would be parallel, whereas a nonzero parallel field has nonzero limiting norm on an asymptotically Euclidean end and cannot be square-integrable. Hence $\xi =0$. Weighted elliptic estimates then give the required asymptotic decay of $\eta$. This supplies elliptic existence of a Witten spinor, rather than assuming it from the sign of a formal integral.

Finally integrate the Witten-Nester identity over $\Sigma$. The ADM boundary term of the Nester two-form isTo identify this boundary term, expand the asymptotic spin connection to first order in $h-\delta$. Its flux is half of $({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})s^i{K}_{\infty }^0$; the extrinsic curvature contribution is $-(k_{ij}-h_{ij}\mathrm{tr}k)s^j{K}_{\infty }^i$. Their coefficients give precisely the two charges displayed above. The derivative of the decaying spinor correction has zero leading integrated flux by antisymmetry, and the higher-order terms vanish with the falloff. ThusChoose a normalized asymptotic spinor field that is an eigenvector with eigenvalue $+1$ of $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$. Then $K_{\infty }^0=1$ and ${\mathbf{K}}_{\infty }=q$. Choosing $q=\mathbf{P}/|\mathbf{P}|$ proves $E\ge |\mathbf{P}|$; if $\mathbf{P}=0$, any such choice gives $E\ge 0$. This is positivity in every asymptotic frame, not only the zero-momentum case.

The zero-energy rigidity also follows from the vanishing integrals. If $E=0$, then $\mathbf{P}=0$ and the construction for a basis of asymptotic spinors produces a basis of parallel fields. Their spinor curvature identity forces the spatially pulled-back spacetime curvature to vanish; the nonnegative matter terms vanish too. The resulting Gauss–Codazzi equations are those of a hypersurface in Minkowski spacetime, and its vacuum development is flat. Completeness and the ordinary Euclidean asymptotic end rule out a nontrivial flat quotient. The initial hypersurface need not have $k=0$: a curved spacelike slice of Minkowski spacetime also has zero ADM energy. The proof works by converting an asymptotic energy charge into a nonnegative bulk integral, with the spinor equation and its existence supplying the crucial bridge.

For a smooth compactly supported $u$, expand the nonnegative squareSince $\nabla \cdot (x/|x{|}^2)=|x{|}^{-2}$ in three dimensions, integration by parts turns the cross term into $-\frac{1}{2}\int |u{|}^2/|x{|}^2$. HenceDensity extends this Hardy inequality in Euclidean space to $H^1({\mathbb{R}}^3)$.

The displayed estimate requires the standard choice $\psi (r)=r$; read literally, the printed ${\psi }^{\prime }(r)=r$ would give $\psi =r^2/2$, $-{\Delta }^2\psi =0$, and would not imply the claimed estimate. For $\psi =|x|$, the distributional bilaplacian of the radial coordinate in three dimensions isThe delta term is nonnegative. The Morawetz action is bounded by $CE$ using Cauchy-Schwarz and the Hardy inequality in Euclidean space. Integrating the identity from $0$ to $T$ and discarding the delta term givesuniformly in $T$, proving the Morawetz estimate for the defocusing wave equation.

A finite-energy stationary solution would make the nonnegative spatial integral on the left constant in time. Its integral over $[0,\infty )$ can be finite only when that spatial integral is zero, so the stationary solution is $u=0$.

Since $w=ru$,and radial integration satisfies $dx=4\pi r^{2}dr$. Multiplying the identity from part 6 by $4\pi$ therefore giveswhich is the modified Morawetz identity.

Choose the constant weight $\psi =1$. The first term on the right vanishes and the second isThe functional on the left is bounded by the conserved energy using the Hardy inequality in Euclidean space. Integrating in time gives another proof of the Morawetz estimate for the defocusing wave equation.