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...


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

  • 1
    $\begingroup$ The inequality in your second Edit can be proved using Moser iteration applied to the elliptic PDE satisfied by $\nabla f$ combined with the L2 bound on $\nabla f$ in terms of the $L^2$ norm of $f$ $\endgroup$ – Deane Yang Aug 28 '12 at 19:44
  • $\begingroup$ @Deane Yang: I don't know much about the Moser iteration - do you have a good reference for a equivalent case (so $f$ no eigenfunction and $\triangle f$ not non-negativ)? Obviously $$(\triangle\nabla f)^j=\nabla^j\triangle f+Ric^{ij}\nabla_if$$ and for $p<\infty$ $$\Vert\nabla f\Vert_{L^p(\mathcal S^2)},\Vert\nabla\nabla f\Vert_{L^2(\mathcal S^2)}\le C_p(\Vert f\Vert_{L^2(\mathcal S^2)}+\Vert\triangle f\Vert_{L^2(\mathcal S^2)}),$$ but then? Or do you know a way to prove a weak inequality -- see last edit. $\endgroup$ – Christopher Nerz Aug 30 '12 at 12:33
  • 1
    $\begingroup$ Unfortunately, I'm terrible with references. But one place where you can find the essential argument is in Appendix C of a paper of mine: Convergence of riemannian manifolds with integral bounds on curvature. II. Annales scientifiques de l'École Normale Supérieure, Sér. 4, 25 no. 2 (1992), p. 179-199. In your case, the Sobolev inequality has a second term in it, but the argument still works. $\endgroup$ – Deane Yang Aug 30 '12 at 13:25
  • $\begingroup$ @Deane Yang: As far as I understand at first sight theorems C.3 and C.7 in this paper of yours, that would need $\Vert\tilde f\Vert_{L^p}$ ($p>n=2$) term on the right-hand side, where $\tilde f$ means the $f$ in the notation of your paper and would correspond to something of second order derivations of $f$ in my notation? So there would again be a $\Vert\nabla\nabla f\Vert_{L^p}$ term on the right hand side? But this I could't use as long as I can't controll them by any terms of $\triangle f$ and $f$ and $\nabla f$ $L^p$-terms ($p\neq\infty$). Or am I missing a central point? $\endgroup$ – Christopher Nerz Aug 30 '12 at 15:46
  • 1
    $\begingroup$ Only $\|\Delta f\|_p$ is needed. $\endgroup$ – Deane Yang Aug 30 '12 at 18:03


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

  • $\begingroup$ Just for completness: Using more technical steps, you can do the last step without using $\Vert f\Vert_{L^2(M)}\le\Vert\triangle f\Vert_{L^2(M)}$. $\endgroup$ – Christopher Nerz Oct 17 '12 at 12:34

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.

  • $\begingroup$ I'm sorry for my poor knowlegde of the latex system in MO. $\endgroup$ – Wang Ming Aug 28 '12 at 15:15
  • $\begingroup$ Thx, but this just works for the laplace on the ball and the euclidean laplace... I need it on the sphere (just the border on the ball) with the induced riemannian metric and the laplace-beltrami operator of this metric... I precised my question. In fact a proof for the "multiplier"-statement on the sphere would be enough $\endgroup$ – Christopher Nerz Aug 28 '12 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.