Compact Laplace-type operators: smooth kernel, closed range and Green inverse

Every $h\in \mathcal{H}$ belongs to $H^2\subset H^1$ and satisfies $Ph=0$; [F1] applied at all integer orders gives a smooth representative and $\mathrm{\parallel }h{\mathrm{\parallel }}_{H^2}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_2$. The kernel is closed in $L^2$: if $h_j\to h$ there, the same estimate for $h_j-h_i$ makes it Cauchy in $H^2$; its $H^2$ limit is $h$ and continuity $P:H^2\to L^2$ gives $Ph=0$. If this closed Hilbert subspace were infinite dimensional, AC permits successively choosing unit vectors orthogonal to the span of earlier choices, producing an orthonormal sequence. The sequence is bounded in $H^2$, so [F1] gives an $L^2$ convergent subsequence, contradicting the distance $\sqrt{2}$ between distinct members. Hence $\mathcal{H}$ is finite dimensional.

There is $c>0$ such that $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le c\mathrm{\parallel }Pu{\mathrm{\parallel }}_2$ for every $u\in H^2\cap {\mathcal{H}}^{\perp }$. Otherwise select $u_j$ in that subspace with $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }P{u}_j{\mathrm{\parallel }}_2<1\mathrm{/}j$. The elliptic estimate bounds it in $H^2$, so compactness gives an $L^2$ convergent subsequence. Applying the elliptic estimate to differences shows this subsequence is Cauchy in $H^2$, since both it and its images under $P$ converge in $L^2$. Its $H^2$ limit $u$ has $Pu=0$, $u\perp \mathcal{H}$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_2=1$, a contradiction. Combining the obtained bound with [F1] yields $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^2}\le C^{\mathrm{\prime }}\mathrm{\parallel }Pu{\mathrm{\parallel }}_2$ on that complement.

Suppose $P{u}_j\to f$ in $L^2$. Subtract the orthogonal projection of each $u_j$ onto the finite smooth space $\mathcal{H}$; its image under $P$ does not change. Step 2.1 makes the resulting sequence Cauchy in $H^2$, so it has a limit $u$ there with $Pu=f$. Thus the range is closed. A vector $y\in L^2$ is orthogonal to the range exactly when $(Pu,y{)}_2=0$ for every $u\in \mathrm{dom}P$, which by the definition of unbounded adjoint says $y\in \mathrm{dom}{P}^{\ast }$ and $P^{\ast }y=0$. Selfadjointness in [F1] identifies this with $y\in \mathcal{H}$. [F2] applied to the closed range therefore gives $\mathrm{ran}P={\mathcal{H}}^{\perp }$ and the claimed $L^2$ decomposition.

For $f\in {\mathcal{H}}^{\perp }$, step 3.1 gives a preimage in $H^2$; subtracting its harmonic projection gives a preimage in $H^2\cap {\mathcal{H}}^{\perp }$, unique because the difference of two lies in $\mathcal{H}\cap {\mathcal{H}}^{\perp }$. Define $Gf$ to be this preimage and define $G=0$ on $\mathcal{H}$. Uniqueness gives linearity. Step 2.1 gives its $L^2\to H^2$ bound. The construction proves both inverse identities and $G\mathrm{\Pi }=\mathrm{\Pi }G=0$. Compactness on $L^2$ follows from [F1]. For $f,g\in {\mathcal{H}}^{\perp }$, symmetry of $P$ gives $(Gf,g{)}_2=(Gf,PGg{)}_2=(PGf,Gg{)}_2=(f,Gg{)}_2$; harmonic cross terms vanish, so $G$ is selfadjoint on all $L^2$.

Choose a finite $L^2$ orthonormal basis $e_1,\dots ,e_N$ of $\mathcal{H}$. Each is smooth, and $\mathrm{\Pi }f={\sum }_j(f,e_j{)}_2{e}_j$. Cauchy–Schwarz gives $\mathrm{\parallel }\mathrm{\Pi }f{\mathrm{\parallel }}_{C^r}\le {\sum }_j\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_2\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_{C^r}$, including the empty-basis case. Thus $\mathrm{\Pi }$ is smoothing and bounded from $L^2$ to every $H^k$. If $f\in H^k$, the equation $PGf=f-\mathrm{\Pi }f$ and [F1] give $Gf\in H^{k+2}$ with $\mathrm{\parallel }Gf{\mathrm{\parallel }}_{H^{k+2}}\le C_k(\mathrm{\parallel }f-\mathrm{\Pi }f{\mathrm{\parallel }}_{H^k}+\mathrm{\parallel }Gf{\mathrm{\parallel }}_2)\le C_k^{\mathrm{\prime }}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^k}$. All integer orders give smooth preservation. The Green identities express every $H^k$ section as $\mathrm{\Pi }f+PGf$ and identify the image summand with $H^k\cap {\mathcal{H}}^{\perp }$. This summand is closed, and both projections are bounded; the smooth version follows and is continuous in every smooth seminorm from the same bounds and Sobolev embedding.

If $P$ is formally nonnegative on smooth sections, density in $H^2$ and continuity of the pairing extend $(Pu,u{)}_2\ge 0$ to all $u\in H^2$. For $f\in L^2$, put $u=Gf$; then $(Gf,f{)}_2=(u,Pu{)}_2\ge 0$, since $u\perp \mathrm{\Pi }f$. This proves the optional positivity claim. The inverse identities of step 4.1 and the smoothing projection of step 5.1 prove the final parametrix assertion.