= Solution
Here is a direct use of the spherical representation that also avoids estimating separate coordinate <derivatives> at the poles. Put $A=\Delta_{S^2}$, $T=\partial_r^2+2r^{-1}\partial_r$, and write $d\omega$ for unit-sphere area. Then $f=Tu+r^{-2}Au$. Expanding its squared <norm> with the physical measure gives
$$
\|f\|_2^2=\|Tu\|_2^2+\|r^{-2}Au\|_2^2+2\int_{1/2}^2\!\int_{S^2}(Tu)(Au)\,d\omega\,dr.
$$
On each boundary sphere $u=0$, so every angular <derivative>, including $Au$ and $\nabla_Su$, is zero there. Radial and angular <integration by parts> consequently give
$$
\int(Tu)(Au)=\int|\nabla_Su_r|^2-\int r^{-2}|\nabla_Su|^2.
$$
Indeed the $u_{rr}$ term gives $\int|\nabla_Su_r|^2$ with boundary remainder $[\int u_rAu]=0$, while the $2u_r/r$ term is $-\int r^{-1}\partial_r|\nabla_Su|^2=-\int r^{-2}|\nabla_Su|^2$, its boundary remainder also being zero. Thus
$$
\|Tu\|_2^2+\|r^{-2}Au\|_2^2+2\int|\nabla_Su_r|^2
=\|f\|_2^2+2\int r^{-2}|\nabla_Su|^2\le C\|f\|_2^2,
$$
where the last step uses the preceding $H^1$ energy estimate and $r\ge1/2$. All unmarked integrals in these two identities are with $dr\,d\omega$.
To control every angular second <derivative>, the <spherical Hessian identity> is
$$
\int_{S^2}|\nabla_S^2w|^2=\int_{S^2}(Aw)^2-\int_{S^2}|\nabla_Sw|^2.
$$
It follows by integrating the derivative-commutation identity $\nabla^a\nabla_a\nabla_bw=\nabla_bAw+\operatorname{Ric}_b{}^c\nabla_cw$ and using $\operatorname{Ric}=h$ on the unit sphere. There is no sphere boundary. Hence the preceding bound controls $\nabla_S^2u$, $\nabla_Su_r$, and $u_{rr}=Tu-2u_r/r$ in their appropriate weighted <norms>.
In an orthonormal polar frame the Cartesian <Hessian matrix> has components
$$
D^2u(e_r,e_r)=u_{rr},\quad D^2u(e_r,e_A)=r^{-1}(\nabla_Su_r)_A-r^{-2}(\nabla_Su)_A,
$$
and $D^2u(e_A,e_B)=r^{-2}(\nabla_S^2u)_{AB}+r^{-1}u_r\delta_{AB}$. Since $r$ is bounded above and below, these bounds plus the first-order estimate prove
$$
\boxed{\|u\|_{H^2(\Omega)}\le C\|f\|_2.}
$$
This is the <direct spherical-shell H2 estimate>. Intrinsic angular integration combines the suggested angular tests, with no spurious boundary at the polar coordinate singularities.
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