$W^{2,p}$ or $W^{1,q}$ regularity for the laplace on a euclidean sphere Hi,
it is easy to prove the $W^{2,2}(\mathcal S^2)$ regularity for the laplace on the (2 dimensional-) standard sphere $\mathcal S^2:=\lbrace x \in\mathbb R^3: \vert x\vert=1 \rbrace\hookrightarrow\mathbb R^3$ by partial integration, getting
 $$\Vert f\Vert_{W^{2,2}(\mathcal S^2)} \le C ( \Vert \triangle f\Vert_{L^2(\mathcal S^2)} + \Vert f\Vert_{L^2(\mathcal S^2)}).$$

But: How do you prove an equivalent $W^{2,p}(\mathcal S^2)$ regularity for $p>2$ in a more or less direct matter? The only interessting part is to prove a

$L^p$-inequality for $D^2f$:
For $p>2$ there exists $1\le q<\infty$ such that
     $$\Vert \nabla(\nabla f) \Vert_{L^p(\mathcal S^2)} \le C_{p,q}( \Vert \triangle f\Vert_{L^p(\mathcal S^2)} + \Vert f\Vert_{W^{1,q}(\mathcal S^2)}).$$

As the rest follows by the sobolev-inequalites and the $W^{2,2}$-case.

It's essential for me to get a direct proof (in particular no contradiction), because I have to adapt it in an approximated sphere on an asymptotic flat Riemannian Manifold.

In fact it would be enough for me to prove
 $$\Vert \nabla(\nabla f) \Vert_{L^p(\mathcal S^2)} \le C_{p,q,r}( \Vert \triangle f\Vert_{L^r(\mathcal S^2)} + \Vert f\Vert_{W^{1,q}(\mathcal S^2)})$$
for some $1\le q<\infty$ and some $1\le r\le\infty$, because of my special situation, but I do not believe that this is true if the inequality above is not.
Thx for some hints...
Elgrimm

Edit: I mean the sphere $\mathcal S^2:=\lbrace x\in\mathbb R^3 : \vert x \vert=1\rbrace$ as riemannian submanifold of $\mathbb R^3$ and the corresponding laplace-beltrami-operator $\triangle$ on $\mathcal S^2$ not the ball as substet of $\mathbb R^3$.

Edit:
For me it would even be sufficient to prove a


$L^\infty$-inequality for $\nabla f$:
There exists $1\le p<\infty$ and $1\le r\le\infty$ such that
    $$ \Vert \nabla f\Vert_{L^\infty(\mathcal S^2)} \le C(\Vert f\Vert_{L^p(\mathcal S^2)} + \Vert\triangle f\Vert_{L^r(\mathcal S^2)}).$$


Which could be concluded from the results above with the sobolev-inequalities.

Edit (sorry for the mass of edits): By looking at $\nabla f$ instead of $f$ it's quite obvious, that it would be a even stronger result to prove the


Weak inequality:
For $p>2$ and $\frac1p+\frac1q=1$ it's true that
    $$ \Vert f\Vert_{W^{1,p}(\mathcal S^2)} \le C(\Vert f\Vert_{L^p(\mathcal S^2)} + \Vert \triangle f\Vert_{W^{-1,q}(\mathcal S^2)}), $$
    where $W^{-1,q}(\mathcal S^2)$ is the dual of $W^{1,q}(\mathcal S^2)$, in particular
    $$\Vert \triangle f\Vert_{W^{-1,q}(\mathcal S^2)}:=\sup_{\Vert\varphi\Vert_{W^{1,q}(\mathcal S^2)}=1}\left\vert\int_{\mathcal S^2}\nabla f\cdot\nabla\varphi\right\vert.$$


 A: Sobolev-Norms:
Let $(M,g)$ be a two-dimensional compact remannian-manifold without boundary $r>0$, define for $f\in\mathcal C^\infty(M)$ the sobolev-inequalites
$$ \Vert f\Vert_{W^{k,p}(M)} := \sum_{l=0}^k r^l\cdot\left\Vert \vert\nabla^l f\vert_g\right\Vert_{L^p(M)}\qquad\forall f\in\mathcal C^1(M),\ k\in\mathbb N_{\ge0}, $$
where $\nabla$ is the levi-civita connection. $W^ {k,p}(M)$ is as usual the completion of $\mathcal C^\infty(M)$ for this norm, and as usual we identifiy elements of $W^{k,p}$-with almost every-where defined functions.

$\triangle$-$W^{1,\infty}$-, -$W^{2,2}$-, -$L^\infty$-regularities:
Let $M$ be a two-dimensional compact manifold without boundary and $c,r\in\mathbb R_{>0}$, such that
$$ \vert M\vert \le cr^2, \quad \Vert \text{Ric}(M) \Vert_{L^\infty(M)} \le \frac c{r^2},\quad \Vert f\Vert_{L^2(M)} \le \frac cr\Vert f\Vert_{W^{1,1}(M)} \quad\forall f\in W^{1,1}(M) $$
and $c>\lambda_1$, where $\lambda_1$ is the smallest positiv eigenvalue of $-\triangle$.
There is a constant $C=C(c)$ such that
$$ \Vert f\Vert_{W^{2,2}(M)} \le C(r^2\Vert \triangle f\Vert_{L^2(M)}+\Vert f\Vert_{L^2(M)}) \qquad\forall f\in W^{2,2}(M), \tag{1} $$
$$ \Vert f\Vert_{L^\infty(M)} \le C(r\Vert \triangle f\Vert_{L^2(M)}+r^{-1}\Vert f\Vert_{L^2(M)}) \qquad\forall f\in W^{2,2}(M), \tag{2} $$
$$ \Vert f\Vert_{W^{1,\infty}(M)} \le Cr^{-\frac2p}(r^2\Vert \triangle f\Vert_{L^p(M)}+\Vert f\Vert_{L^p(M)}) \qquad\forall f\in W^{2,p}(M),\ p>2, \tag{3} $$
if $r>C$.

Proof [as Deane Yang suggested by a Moser-iteration]:
Be setting $g:=\vert f\vert^{\frac 12}$, we get the usual sobolev-inequalites
$$\Vert f\Vert_{L^q(M)} \le \frac cr\Vert f\Vert_{W^{1,p}(M)} \quad\forall f\in W^{1,p}(M),\ 1\le p< 2: \ q=\frac{2p}{2-p}. \tag 4$$
By an argument as as in [1] we conclude the additional sobolev-inequality
$$\Vert f\Vert_{L^\infty(M)} \le cr^{-\frac2p}\Vert f\Vert_{W^{1,p}(M)} \qquad\forall f\in W^{1,p}(M), 2< p. \tag 5 $$
Using that $div\nabla X=g^{ij}Ric(X,e_i)e_j+\nabla div X$ for any frame $\lbrace e_1,e_2\rbrace$ and any smooth vector field $X$, we conclude $(1)$ by defining $X=\nabla f$ and partial integration and therefore we can deduce $(2)$ with $(5)$.
Let $f\in W^{2,p}(M)$ be such that $\int f=0$, in particular $\Vert f\Vert_{L^2(M)} \le c \Vert\triangle f\Vert_{L^2(M)}$, and assume there exists a polnom $P$ and a constant $C$ such that for $q=\frac{2p}{p-2}$
$$ \Vert \vert\nabla f\vert^{2s}\Vert_{L^p(M)} \le P(s)\left(\left(\frac Csr^{\frac2p}\Vert\triangle f\Vert_{L^q(M)}\right)^{2r}+r^{-\frac2p}\Vert\vert\nabla f\vert^r\Vert_{L^p(M)}\right). \tag6 $$
Setting $a_i:=(r^{2^i-\frac2p}\Vert\vert\nabla f\vert^{2^i}\Vert_{L^p(M)})^{2^{-i}}$, we conclude by basic analysis
$$ r\Vert \nabla f\Vert_{L^\infty(M)} \le C\left(r^{1-\frac2p}\Vert\triangle f\Vert_{L^p(M)}+r^{1+\frac2p}\Vert\nabla f\Vert_{L^{\frac{2p}{2-p}}(M)}\right). $$
This inequality obviously also holds for $f+d$ ($d\in\mathbb R)$. So we only have to prove $(6)$ for a polnyom and a constant not depending on $f\in\mathcal C^\infty(M)$ with $\int f\ d\mu=0$. By basic analysis and some partial integrations we see
$$ \int \vert\nabla(\vert\nabla f\vert^s)\vert^2\ d\mu \le -s\int\vert\nabla f\vert^{2s-2}g(\nabla f,\triangle\nabla f). $$
We conclude by hölder-inequalities and partial integration
$$ \int \vert\nabla(\vert\nabla f\vert^s)\vert^2\ d\mu \le Csr^{\frac2{ps}}\Vert \vert\nabla f\vert^{2s}\Vert_{L^p(M)}^{\frac{s-1}s}\underbrace{\left(\begin{aligned}\Vert\text{Ric}\Vert_{L^\infty(M)}\cdot\Vert\nabla f\Vert_{L^{\frac{2p}{p-1}}(M)}^2+\Vert\triangle f\Vert_{L^{\frac{2p}{p-1}}(M)}^2 \\\\ + \Vert\triangle f\Vert_{L^2(M)}\cdot\Vert\triangle f\Vert_{L^{\frac{2p}{p-2}}(M)} \\\\ + \Vert\text{ric}\Vert_{L^\infty(M)}^{\frac12}\Vert\nabla f\Vert_{L^2(M)}\cdot\Vert\triangle f\Vert_{L^{\frac{2p}{p-2}}(M)}\end{aligned}\right)}_{=: c_f}. $$
By some Yang- and Hölderinequalities and using $\Vert f\Vert_{L^2(M)}\le c\Vert \triangle f\Vert_{L^2(M)}$ we see for $q:=\frac{2p}{2-p}$ and the inequality for $\text{ric}$
$$ c_f \le C_pr^{\frac2p}\Vert\triangle f\Vert_{L^q(M)}^2. $$
By the Sobolev- and Young-inequalities we therefore can conclude $(6)$ for $P(s)=s^6$ using the inequality for $\text{ric}$.

[1]: Stampacchia, Guido: Régularisation des solutions de problèmes aux limites
elliptiques à données discontinues. In: Proceedings of the International Sympo-
sium on Linear Spaces. Hebrew University of Jerusalem : Jerusalem Acad. Pr.,
July 1960 (A publication of the Israel Academy of Sciences and Humanities),
S. 399–408
A: Let $\Omega \subset R^2$ is a ball. Consider the equation
$$
-\triangle u = f(x), \quad x \in \Omega 
$$
$$
u \big|_{\partial \Omega} = 0.
$$
It suffices to prove that for $p \geq 2$
$$\|D^2u\|_{L^p(\Omega)} \leq C \|f\|_{L^p(\Omega)}.
$$
At first, as you know, using integration by parts we have
$$
\|u\|_{H^1(\Omega)} \leq C \|f\|_{L^2} \leq C \|f\|_{L^p(\Omega)}.
$$
Then consider a cutoff function $\eta \in C^\infty_0(\Omega)$, denote by $v = \eta u$, then $v$ satisfies the equation
$$
-\triangle v = \eta f - 2\nabla u \cdot \nabla \eta - \triangle \eta u, \quad x \in R^2.
$$
It's known that $\xi_i\xi_j/|\xi|^2$ is an $L^{q}$ multiplier, that is,
$$
\|\partial_i\partial_j u\|_{L^q(R^2)}  \leq C \|\triangle u\|_{L^q(R^2)}, \quad q \in (1, \infty).
$$
Using the above facts, notice the support of $\eta$ we obtain
$$
\|u\|_{H^2{(\Omega)}} \leq C \|f\|_{L^p}(\Omega).
$$
Then the Sobolev embedding theorem yields that
$$
\|u\|_{W^{1,q}(\Omega)} \leq C \|f\|_{L^p(\Omega)}, \quad 2 \leq q < \infty.
$$
Proceed the above argument again, we find
$$
\|D^2u\|_{L^p(\Omega)} \leq C \|f\|_{L^p(\Omega)}
$$
as desired.
