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 $$\Vert \triangle f\Vert_{W^{-1,q}(\mathcal S^2)}:=\sup_{\Vert\varphi\Vert_{W^{1,q}(\mathcal S^2)}=1}\left\vert\int_{\mathcal S^2}\nabla f\cdot\nabla\varphi\right\vert.$$

4more comments