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
    Commented Aug 28, 2012 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. $\endgroup$ Commented Aug 30, 2012 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
    Commented Aug 30, 2012 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? $\endgroup$ Commented Aug 30, 2012 at 15:46
  • 1
    $\begingroup$ Only $\|\Delta f\|_p$ is needed. $\endgroup$
    – Deane Yang
    Commented Aug 30, 2012 at 18:03

2 Answers 2



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 $\Vert f\Vert_{L^2(M)}\le\Vert\triangle f\Vert_{L^2(M)}$. $\endgroup$ Commented Oct 17, 2012 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.

  • $\begingroup$ 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 $\endgroup$ Commented Aug 28, 2012 at 15:56

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