11.24 Fredholm theory for a continuous Hermitian formDB480BD29A98

Proof. Fix a smooth positive Hermitian form $\eta $ and put $L_ s=(1-s)L_\eta +sL_\sigma $, $0\leq s\leq 1$. On a finite holomorphic cover these operators have uniformly positive real principal matrices and a common modulus of continuity. They have no lower-order terms in these coordinates. Write $d=2n$ for the real dimension.

Local estimates. For a constant positive real matrix $a_0$, the symbols $\xi _ i\xi _ j/(1+a_0^{kl}\xi _ k\xi _ l)$ satisfy the order-zero multiplier estimates. Consequently

Freeze the principal matrix of $L_ s$ at a point and shrink the ball until the coefficient oscillation times this constant is less than $1/2$. The oscillation term can then be absorbed on the left. Differentiating a cutoff adds first derivatives; the interpolation inequality $\| Dz\| _ p\leq \varepsilon \| D^2z\| _ p+C_\varepsilon \| z\| _ p$ absorbs them. This inequality follows by comparing $Dz$ with its mollification at scale $\varepsilon $ and estimating the derivative of the mollified function by $C\varepsilon ^{-1}\| z\| _ p$. A partition of unity and smooth density give, uniformly in $s$,

Maximum principles. We give the local estimate needed to pass comparison to $W^{2,p}$. Let $L=a^{ij}D_{ij}$ on a bounded smooth coordinate domain $D$, with $a\geq \vartheta I$, and let $z$ be smooth with nonnegative boundary values and minimum $-M<0$. Set $R=\operatorname {diam}D$. Every vector $\xi $ with $|\xi |<M/(2R)$ is the gradient at an interior lower supporting point of $z$: minimize $z(x)-\xi \cdot (x-x_0)$, where $z(x_0)=-M$. Its boundary values exceed $-M/2$, whereas its value at $x_0$ is $-M$. At a supporting point $D^2z\geq 0$, and the arithmetic–geometric mean inequality gives

The area formula for the gradient, restricted to these supporting points, therefore bounds the volume of the slope ball by $(d\vartheta )^{-d}\int _ D((Lz)^+)^ d$. Thus

This calculation uses no derivative of $a$. Approximate a $W^{2,p}$ function smoothly on the closure of a smaller smooth domain. The approximation converges in $C^1$, and its $L$ errors converge in $L^ d$ because $p>d$. Correct the vanishing boundary error in (11.24.3). Passage to the limit shows that $Lv\leq 0$ and $v|_{\partial D}\geq 0$ imply $v\geq 0$.

The same comparison proves the strong maximum principle. If $v\in W^{2,p}$, $Lv\geq 0$, and $v$ has an interior maximum $M$, put $w=M-v\geq 0$. If $w$ is positive somewhere and zero somewhere, a small ball contained in $\{ w>0\} $ and tangent to its zero set can be chosen inside a coordinate chart. On the outer half-annulus of this ball,

is positive and has $L\zeta >0$ for sufficiently large $k$, by uniform ellipticity. A small positive multiple of $\zeta $ is bounded above by $w$ on both boundary spheres, hence throughout the annulus by comparison. At the tangent point this forces a positive inward derivative of $w$, contradicting $Dw=0$. Connectedness proves that $v$ is constant. Applied to $L_ s$ on $X$, it gives $\ker L_ s=\mathbb R$. If $(L_ s-1)v=0$ and $v$ has a positive maximum, apply the strong maximum principle on its positive component, where $L_ sv=v>0$. Constancy contradicts that equation. Applying the argument to $-v$ rules out a negative minimum. Hence $\ker (L_ s-1)=0$.

Index and range. Compactness of $W^{2,p}\hookrightarrow L^ p$ and (11.24.2) now give

uniformly in $s$. Otherwise normalize a contradicting sequence in $W^{2,p}$, pass to $s_ j\to s$ and a strongly $L^ p$ convergent subsequence, and apply (11.24.2) to differences. Operator norm continuity in $s$ gives strong $W^{2,p}$ convergence to a nonzero member of $\ker (L_ s-1)$, which is impossible. The smooth operator $L_\eta -1$ has index zero and trivial kernel, hence is an isomorphism, by smooth elliptic theory. The method of continuity and (11.24.4) show that every $L_ s-1$ is an isomorphism. Finally, $L_\sigma $ differs from $L_\sigma -1$ by the compact inclusion $W^{2,p}\hookrightarrow L^ p$, so it is Fredholm of index zero. Its one-dimensional kernel makes its range have codimension one. $L^ p$ duality gives the annihilator assertion. $\square$