The LIR Method. $L^{r}$ Solutions of Elliptic Equation in a Complete Riemannian Manifold

Abstract

We introduce the Local Increasing Regularity Method (LIRM) which allows us to get from local a priori estimates, on solutions u of a linear equation $\displaystyle Du=\omega ,$global ones. As an application we shall prove that if D is an elliptic linear differential operator of order m with ${\mathcal {C}}^{\infty }$ coefficients operating on the sections of a complex vector bundle $\displaystyle G:=(H,\pi ,M)$ over a compact Riemannian manifold M without boundary and $\omega \in L^{r}_{G}(M)\cap (\mathrm {k}\mathrm {e}\mathrm {r}D^{*})^{\perp },$ then there is a $u\in W^{m,r}_{G}(M)$ such that $Du=\omega$ on M. Next we investigate the case of a compact manifold with boundary by using the “Riemannian double manifold.” In the last sections we study the more delicate case of a complete but non-compact Riemannian manifold by the use of adapted weights.

Appendix

We shall use the following lemma.

Lemma 7.1

Let (M, g) be a Riemannian manifold; then with $\displaystyle R(x)=R_{\epsilon }(x)=$ the $\epsilon$ admissible radius at $\displaystyle x\in M$ and $\displaystyle d(x,y)$ the Riemannian distance on $\displaystyle (M,g)$ we get

$$\begin{aligned} \displaystyle d(x,y)\le \frac{1}{4}(R(x)+R(y))\Rightarrow R(x)\le 4R(y). \end{aligned}$$

Proof

Let $\displaystyle x,y\in M::d(x,y)\le \frac{1}{4}(R(x)+R(y))$ and suppose for instance that $\displaystyle R(x)\ge R(y).$ Then $\displaystyle y\in B(x,R(x)/2)$ and hence we have $\displaystyle B(y,R(x)/4)\subset B(x,\frac{3}{4}R(x)).$ But by the definition of $\displaystyle R(x),$ the ball $\displaystyle B(x,\frac{3}{4}R(x))$ is admissible and this implies that the ball $\displaystyle B(y,R(x)/4)$ is also admissible for exactly the same constants and the same chart; this implies that $\displaystyle R(y)\ge R(x)/4.$$\square$

1.1 Vitali Covering 

Lemma 7.2

Let ${\mathcal {F}}$ be a collection of balls $\lbrace B(x,r(x))\rbrace$ in a metric space, with $\forall B(x,r(x))\in {\mathcal {F}},\ 0<r(x)\le R.$ There exists a disjoint subcollection ${\mathcal {G}}$ of ${\mathcal {F}}$ with the following property: every ball B in ${\mathcal {F}}$ intersects a ball C in ${\mathcal {G}}$ and $\displaystyle B\subset 5C.$

This is a well-known lemma, see for instance  [13], Section 1.5.1.

Fix $\epsilon >0$ and let $\displaystyle \forall x\in M,\ r(x):=R_{\epsilon }(x)/120,\$where $\displaystyle R_{\epsilon }(x)$ is the admissible radius at $\displaystyle x,$ and we built a Vitali covering with the collection ${\mathcal {F}}:=\lbrace B(x,r(x))\rbrace _{x\in M}.$ The previous lemma gives a disjoint subcollection ${\mathcal {G}}$ such that every ball B in ${\mathcal {F}}$ intersects a ball C in ${\mathcal {G}}$ and we have $\displaystyle B\subset 5C.$ We set ${\mathcal {G}}^{\prime }:=\lbrace x_{j}\in M::B(x_{j},r(x_{j}))\in {\mathcal {G}}\rbrace$ and ${\mathcal {C}}_{\epsilon }:=\lbrace B(x,5r(x)),\ x\in {\mathcal {G}}^{\prime }\rbrace .$ We shall call ${\mathcal {C}}_{\epsilon }$ the $m,\epsilon$admissible covering of $\displaystyle (M,g).$

We shall fix $m\ge 2$ and we omit it in order to ease the notation.

Recall that $\epsilon <1,$ then we have:

Proposition 7.3

Let (M, g) be a Riemannian manifold. The overlap of the $\epsilon$ admissible covering ${\mathcal {C}}_{\epsilon }$ is less than $\displaystyle T=\frac{(1+\epsilon )^{n/2}}{(1-\epsilon )^{n/2}}(120)^{n},$ i.e.,

$$\begin{aligned} \forall x\in M,\ x\in B(y,5r(y)) \end{aligned}$$

for at most T such balls, where $B(y,r(y))\in {\mathcal {G}}.$

So we have

$$\begin{aligned} \forall f\in L^{1}(M),\ \sum _{j\in {\mathbb {N}}}{\int _{B_{j}}{\left| {f(x)}\right| dv_{g}(x)}}\le T{\left\| {f}\right\| }_{L^{1}(M)}. \end{aligned}$$

Proof

Let $B_{j}:=B(x_{j},r(x_{j}))\in {\mathcal {G}}$ and suppose that $\displaystyle x\in \bigcap _{j=1}^{k}{B(x_{j},5r(x_{j}))}.$ Then we have

$$\begin{aligned} \displaystyle \forall j=1,\ldots ,k,\ d(x,x_{j})\le 5r(x_{j}) \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} \displaystyle d(x_{j},x_{l})&\le d(x_{j},x)+d(x,x_{l})\le 5(r(x_{j})+r(x_{l}))\le \frac{1}{4}(R(x_{j})+R(x_{l}))\\&\Rightarrow R(x_{j})\le 4R(x_{l}) \end{aligned} \end{aligned}$$

and by exchanging $\displaystyle x_{j}$ and $\displaystyle x_{l},\ R(x_{l})\le 4R(x_{j}).$

So we get

$$\begin{aligned} \displaystyle \forall j,l=1,\ldots ,k,\ r(x_{j})\le 4r(x_{l}),\ r(x_{l})\le 4r(x_{j}). \end{aligned}$$

Now the ball $\displaystyle B(x_{j},5r(x_{j})+5r(x_{l}))$ contains $\displaystyle x_{l}$ and hence the ball $\displaystyle B(x_{j},5r(x_{j})+6r(x_{l}))$ contains the ball $\displaystyle B(x_{l},r(x_{l})).$ But, because $\displaystyle r(x_{l})\le 4r(x_{j}),$ we get

$$\begin{aligned} \displaystyle B(x_{j},5r(x_{j})+6{\times }4r(x_{j}))=B(x_{j},r(x_{j})(5+24))\supset B(x_{l},r(x_{l})). \end{aligned}$$

The balls in ${\mathcal {G}}$ being disjoint, we get, setting $\displaystyle B_{l}:=B(x_{l},\ r(x_{l})),$

$$\begin{aligned} \displaystyle \ \sum _{j=1}^{k}{\mathrm {V}\mathrm {o}\mathrm {l}(B_{l})}\le \mathrm {V}\mathrm {o}\mathrm {l}(B(x_{j},29r(x_{j}))). \end{aligned}$$

The Lebesgue measure read in the chart $\varphi$ and the canonical measure $dv_{g}$ on $\displaystyle B(x,R_{\epsilon }(x))$ are equivalent; precisely because of condition (1) in the admissible ball definition, we get that

$$\begin{aligned} \displaystyle (1-\epsilon )^{n}\le \left| {\mathrm {d}\mathrm {e}\mathrm {t}g}\right| \le (1+\epsilon )^{n}, \end{aligned}$$

and the measure $dv_{g}$ read in the chart $\varphi$ is $dv_{g}={\sqrt{\left| {\mathrm {d}\mathrm {e}\mathrm {t}g_{ij}}\right| }}d\xi ,$ where $\displaystyle d\xi$ is the Lebesgue measure in ${\mathbb {R}}^{n}.$ In particular,

$$\begin{aligned} \displaystyle \forall x\in M,\ \mathrm {V}\mathrm {o}\mathrm {l}(B(x,\ R_{\epsilon }(x)))\le (1+\epsilon )^{n/2}\nu _{n}R^{n}, \end{aligned}$$

where $\nu _{n}$ is the euclidean volume of the unit ball in ${\mathbb {R}}^{n}.$

Now because $\displaystyle R(x_{j})$ is the admissible radius and $\displaystyle 4{\times }29r(x_{j})<R(x_{j}),$ we have

$$\begin{aligned} \displaystyle \mathrm {V}\mathrm {o}\mathrm {l}(B(x_{j},29r(x_{j})))\le 29^{n}(1+\epsilon )^{n/2}v_{n}r(x_{j})^{n}. \end{aligned}$$

On the other hand we also have

$$\begin{aligned} \displaystyle \mathrm {V}\mathrm {o}\mathrm {l}(B_{l})\ge v_{n}(1-\epsilon )^{n/2}r(x_{l})^{n}\ge v_{n}(1-\epsilon )^{n/2}4^{-n}r(x_{j})^{n}, \end{aligned}$$

and hence

$$\begin{aligned} \displaystyle \ \sum _{j=1}^{k}{(1-\epsilon )^{n/2}4^{-n}r(x_{j})^{n}}\le 29^{n}(1+\epsilon )^{n/2}r(x_{j})^{n}, \end{aligned}$$

so finally

$$\begin{aligned} \displaystyle k\le (29{\times }4)^{n}\frac{(1+\epsilon )^{n/2}}{(1-\epsilon )^{n/2}}, \end{aligned}$$

which means that $\displaystyle T\le \frac{(1+\epsilon )^{n/2}}{(1-\epsilon )^{n/2}}(120)^{n}.$

Saying that any $\displaystyle x\in M$ belongs to at most T balls of the covering $\displaystyle \lbrace B_{j}\rbrace$ means that ${\sum _{j\in {\mathbb {N}}}{{\mathbb {1}}_{B_{j}}(x)}\le T}$, and this implies easily that

$$\begin{aligned} \displaystyle \forall f\in L^{1}(M),\ \sum _{j\in {\mathbb {N}}}{\int _{B_{j}}{\left| {f(x)}\right| dv_{g}(x)}}\le T{\left\| {f}\right\| }_{L^{1}(M)}. \end{aligned}$$

$\square$

1.2 Sobolev Spaces

We have to define the Sobolev spaces in our setting, following Hebey [17], p. 10.

First define the covariant derivatives by $\displaystyle (\nabla u)_{j}:=\partial _{j}u$ in local coordinates, while the components of $\nabla ^{2}u$ are given by

$$\begin{aligned} (\nabla ^{2}u)_{ij}=\partial _{ij}u-\Gamma ^{k}_{ij}\partial _{k}u, \end{aligned}$$

(7.1)

with the convention that we sum over repeated index. The Christoffel $\displaystyle \Gamma ^{k}_{ij}$ verify [7]:

$$\begin{aligned} \Gamma ^{k}_{ij}=\frac{1}{2}g^{il}\left( \frac{\partial g_{kl}}{\partial x^{j}}+\frac{\partial g_{lj}}{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right) . \end{aligned}$$

(7.2)

If $\displaystyle k\in {\mathbb {N}}$ and $r\ge 1$ are given, we denote by ${\mathcal {C}}^{r}_{k}(M)$ the space of smooth functions $u\in {\mathcal {C}}^{\infty }(M)$ such that $\displaystyle \ \left| {\nabla ^{j}u}\right| \in L^{r}(M)$ for $\displaystyle j=0,\ldots ,k.$ Hence

$$\begin{aligned} \displaystyle {\mathcal {C}}^{r}_{k}(M):=\left\{ u\in {\mathcal {C}}^{\infty }(M),\ \forall j=0,\ldots ,k,\ \int _{M}{\left| {\nabla ^{j}u}\right| ^{r}dv_{g}}<\infty \right\} . \end{aligned}$$

Now we have [17].

Definition 7.4

The Sobolev space $\displaystyle W^{k,r}(M)$ is the completion of ${\mathcal {C}}^{r}_{k}(M)$ with respect to the norm:

$$\begin{aligned} {\left\| {u}\right\| }_{W^{k,r}(M)}=\sum _{j=0}^{k}{{\left( {\int _{M}{\left| {\nabla ^{j}u}\right| ^{r}dv_{g}}}\right) }^{1/r}}. \end{aligned}$$

We extend in a natural way this definition to the case of G-forms.

Let the Sobolev exponents $\displaystyle S_{k}(r)$ be as in Definition 1.7, then the k th Sobolev embedding is true if we have

$$\begin{aligned} \displaystyle \forall u\in W^{k,r}(M),\ u\in L^{S_{k}(r)}(M). \end{aligned}$$

This is the case in ${\mathbb {R}}^{n},$ or if M is compact, or if M has a Ricci curvature bounded from below and $\displaystyle \inf \ _{x\in M}v_{g}(B_{x}(1))\ge \delta >0,$ due to Varopoulos [27], see Theorem 3.14, p. 31 in [17].

Lemma 7.5

We have the Sobolev comparison estimates where $\displaystyle B(x,R)$ is a $\epsilon$ admissible ball in M and $\varphi \ ,\ B(x,R)\rightarrow {\mathbb {R}}^{n}$ is the admissible chart relative to $\displaystyle B(x,R),$

$$\begin{aligned} \displaystyle \forall u\in W^{m,r}(B(x,R)),\ {\left\| {u}\right\| }_{W^{m,r}(B(x,R))}\le (1+\epsilon C){\left\| {u\circ \varphi ^{-1}}\right\| }_{W^{m,r}(\varphi (B(x,R)))}, \end{aligned}$$

and, with $\displaystyle B_{e}(0,t)$ the euclidean ball in ${\mathbb {R}}^{n}$ centered at 0 and of radius $\displaystyle t,$

$$\begin{aligned} \displaystyle \ {\left\| {v}\right\| }_{W^{m,r}(B_{e}(0,(1-\epsilon )R))}\le (1+2C\epsilon ){\left\| {u}\right\| }_{W^{m,r}(B(x,R))}. \end{aligned}$$

Proof

We have to compare the norms of $\displaystyle u,\ \nabla u,...,\ \nabla ^{m}u$ with the corresponding ones for $\displaystyle v:=u\circ \varphi ^{-1}$ in ${\mathbb {R}}^{n}.$

First we have because $\displaystyle (1-\epsilon )\delta _{ij}\le g_{ij}\le (1+\epsilon )\delta _{ij}$ in $\displaystyle B(x,R)$:

$$\begin{aligned} \displaystyle B_{e}(0,(1-\epsilon )R)\subset \varphi (B(x,R))\subset B_{e}(0,(1+\epsilon )R). \end{aligned}$$

Because

$$\begin{aligned} \displaystyle \ \sum _{\left| {\beta }\right| \le m-1}{\sup \ _{i,j=1,\ldots ,n,\ y\in B_{x}(R)}\left| {\partial ^{\beta }g_{ij}(y)}\right| }\le \epsilon \hbox { in } \displaystyle B(x,R), \end{aligned}$$

we have the estimates, with $\displaystyle \forall y\in B(x,R),\ z:=\varphi (y),$

$$\begin{aligned} \displaystyle \forall y\in B(x,R),\ \left| {u(y)}\right| =\left| {v(z)}\right| ,\ \ \left| {\nabla u(y)}\right| \le (1+C\epsilon )\left| {\partial v(z)}\right| . \end{aligned}$$

Because of (7.2) and (7.1) we get

$$\begin{aligned} \displaystyle \forall y\in B(x,R),\ \left| {\nabla ^{2}u(y)}\right| \le \left| {\partial ^{2}v(z)}\right| +\epsilon C\left| {\partial v(z)}\right| . \end{aligned}$$

And taking more derivatives, because

$$\begin{aligned} \displaystyle \sum _{\left| {\beta }\right| \le m-1}{\sup \ _{i,j=1,\ldots ,n,\ y\in B_{x}(R)}\left| {\partial ^{\beta }g_{ij}(y)}\right| }\le \epsilon , \end{aligned}$$

we get, for $2\le k\le m,$

$$\begin{aligned} \displaystyle \forall y\in B(x,R),\ \left| {\nabla ^{k}u(y)}\right| \le \left| {\partial ^{k}v(z)}\right| +\epsilon (C_{1}\left| {\partial v(z)}\right| +\cdot \cdot \cdot +C_{k-1}\left| {\partial ^{k-1}v(z)}\right| ). \end{aligned}$$

Integrating this we get for $2\le k\le m,$

$$\begin{aligned} \displaystyle {\left\| {\nabla ^{k}u}\right\| }_{L^{r}(B(x,R))}\le & {} {\left\| {\left| {\partial ^{k}v}\right| +\epsilon (C_{1}\left| {\partial v(z)}\right| +\cdot \cdot \cdot +C_{k-1}\left| {\partial ^{k-1}v(z)}\right| )}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}\\ \displaystyle\le & {} {\left\| {\partial ^{k}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+C_{1}\epsilon {\left\| {\partial v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+\cdot \cdot \cdot \\&+\,C_{k-1}\epsilon {\left\| {\partial ^{k-1}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}, \end{aligned}$$

and

$$\begin{aligned} \displaystyle \ {\left\| {\nabla u}\right\| }_{L^{r}(B(x,R))}\le (1+C\epsilon ){\left\| {\partial v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}. \end{aligned}$$

We also have the reverse estimates

$$\begin{aligned} \displaystyle \ {\left\| {\partial ^{k}v}\right\| }_{L^{r}(B_{e}(0,(1-\epsilon )R))}\le & {} {\left\| {\nabla ^{k}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+C_{1}\epsilon {\left\| {\nabla v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+\cdot \cdot \cdot \\&\displaystyle +\,C_{k-1}\epsilon {\left\| {\nabla ^{k-1}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}, \end{aligned}$$

and

$$\begin{aligned} \displaystyle \ {\left\| {\partial v}\right\| }_{L^{r}(B_{e}(0,(1-\epsilon )R))}\le (1+C\epsilon ){\left\| {\nabla u}\right\| }_{L^{r}(B(x,R))}. \end{aligned}$$

So, using that

$$\begin{aligned} \displaystyle \ {\left\| {u}\right\| }_{W^{k,r}(B(x,R))}={\left\| {\nabla ^{k}u}\right\| }_{L^{r}(B(x,R))}+\cdot \cdot \cdot +{\left\| {\nabla u}\right\| }_{L^{r}(B(x,R))}+{\left\| {u}\right\| }_{L^{r}(B(x,R))}, \end{aligned}$$

we get

$$\begin{aligned} \displaystyle \ {\left\| {u}\right\| }_{W^{k,r}(B(x,R))}^{r} &\!\le \!{\left\| {\partial ^{k}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+C_{2}\epsilon {\left\| {\partial ^{2}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}+\cdot \cdot \cdot \\&\displaystyle +\quad C_{k-1}\epsilon {\left\| {\partial ^{k-1}v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}\!+\!(1+C\epsilon ){\left\| {\partial v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}\\&\displaystyle +\quad {\left\| {v}\right\| }_{L^{r}(B_{e}(0,(1+\epsilon )R))}\\ \displaystyle\le & {} (1+2\epsilon C){\left\| {v}\right\| }_{W^{k,r}(B_{e}(0,(1+\epsilon )R))}. \end{aligned}$$

Again all these estimates can be reversed so we also have

$$\begin{aligned} \displaystyle \ {\left\| {v}\right\| }_{W^{m,r}(B_{e}(0,(1-\epsilon )R))}\le (1+2C\epsilon ){\left\| {u}\right\| }_{W^{m,r}(B(x,R))}. \end{aligned}$$

This ends the proof of the lemma. $\square$

We have to study the behavior of the Sobolev embeddings w.r.t. the radius. Set $\displaystyle B_{R}:=B_{e}(0,R).$

Lemma 7.6

We have, with $\displaystyle t=S_{m}(r),$

$$\begin{aligned} \displaystyle \forall R,\ 0<R\le 1,\ \forall u\in W^{m,r}(B_{R}),\ {\left\| {u}\right\| }_{L^{t}(B_{R})}\le CR^{-m}\ {\left\| {u}\right\| }_{W^{m,r}(B_{R})} \end{aligned}$$

the constant C depending only on $\displaystyle n,\ r.$

Proof

Start with $\displaystyle R=1,$ and then we have by Sobolev embeddings with $\displaystyle t=S_{m}(r),$

$$\begin{aligned} \forall v\in W^{m,r}(B_{1}),\ {\left\| {v}\right\| }_{L^{t}(B_{1})}\le C{\left\| {v}\right\| }_{W^{m,r}(B_{1})}, \end{aligned}$$

(7.3)

where $\displaystyle C$ depends only on n and $\displaystyle r.$ For $\displaystyle u\in W^{m,r}(B_{R})$ we set

$$\begin{aligned} \displaystyle \forall x\in B_{1},\ y:=Rx\in B_{R},\ v(x):=u(y). \end{aligned}$$

Then we have

$$\begin{aligned} \displaystyle \partial v(x)= & {} \partial u(y){\times }\frac{\partial y}{\partial x}=R\partial u(y);\\ \displaystyle \partial ^{2}v(x)= & {} \partial ^{2}u(y){\times }\left( \frac{\partial y}{\partial x}\right) ^{2}=R^{2}\partial ^{2}u(y);\ ...\ ;\\ \displaystyle \partial ^{m}v(x)= & {} \partial ^{m}u(y){\times }\left( \frac{\partial y}{\partial x}\right) ^{m}=R^{m}\partial ^{m}u(y). \end{aligned}$$

So we get, because the Jacobian for this change of variables is $\displaystyle R^{-n},$

$$\begin{aligned} \displaystyle \ {\left\| {\partial v}\right\| }_{L^{r}(B_{1})}^{r}=\int _{B_{1}}{\left| {\partial v(x)}\right| ^{r}dm(x)}=\int _{B_{R}}{\left| {\partial u(y)}\right| ^{r}\frac{R^{r}}{R^{n}}dm(x)}=R^{r-n}{\left\| {\partial u}\right\| }_{L^{r}(B_{R})}^{r}. \end{aligned}$$

So

$$\begin{aligned} {\left\| {\partial u}\right\| }_{L^{r}(B_{R})}=R^{-1+n/r}{\left\| {\partial v}\right\| }_{L^{r}(B_{1})}. \end{aligned}$$

(7.4)

The same way we get

$$\begin{aligned} {\left\| {\partial ^{m}u}\right\| }_{L^{r}(B_{R})}=R^{-m+n/r}{\left\| {\partial ^{m}v}\right\| }_{L^{r}(B_{1})} \end{aligned}$$

(7.5)

and of course $\displaystyle \ {\left\| {u}\right\| }_{L^{r}(B_{R})}=R^{n/r}{\left\| {v}\right\| }_{L^{r}(B_{1})}.$

So with 7.3 we get

$$\begin{aligned} {\left\| {u}\right\| }_{L^{t}(B_{R})}=R^{n/t}{\left\| {v}\right\| }_{L^{t}(B_{1})}\le CR^{n/t}{\left\| {v}\right\| }_{W^{m,r}(B_{1})}. \end{aligned}$$

(7.6)

But

$$\begin{aligned} \displaystyle \ {\left\| {u}\right\| }_{W^{m,r}(B_{R})}:={\left\| {u}\right\| }_{L^{r}(B_{R})}+{\left\| {\partial u}\right\| }_{L^{r}(B_{R})}+\cdot \cdot \cdot +{\left\| {\partial ^{m}u}\right\| }_{L^{r}(B_{R})}, \end{aligned}$$

and

$$\begin{aligned} \displaystyle \ {\left\| {v}\right\| }_{W^{m,r}(B_{1})}:={\left\| {v}\right\| }_{L^{r}(B_{1})}+{\left\| {\partial v}\right\| }_{L^{r}(B_{1})}+\cdot \cdot \cdot +{\left\| {\partial ^{m}v}\right\| }_{L^{r}(B_{1})}, \end{aligned}$$

so

$$\begin{aligned} \displaystyle \ {\left\| {v}\right\| }_{W^{m,r}(B_{1})}:=R^{-n/r}{\left\| {u}\right\| }_{L^{r}(B_{R})}+R^{1-n/r}{\left\| {\partial u}\right\| }_{L^{r}(B_{R})}+\cdot \cdot \cdot +R^{m-n/r}{\left\| {\partial ^{m}u}\right\| }_{L^{r}(B_{R})}. \end{aligned}$$

Because we have $\displaystyle R\le 1,$ we get

$$\begin{aligned} \displaystyle \ {\left\| {v}\right\| }_{W^{m,r}(B_{1})}\le & {} R^{-n/r}\big ({\left\| {u}\right\| }_{L^{r}(B_{R})}+{\left\| {\partial u}\right\| }_{L^{r}(B_{R})}+\cdot \cdot \cdot +{\left\| {\partial ^{m}u}\right\| }_{L^{r}(B_{R})}\big )\\= & {} R^{-n/r}{\left\| {u}\right\| }_{W^{m,r}(B_{R})}. \end{aligned}$$

Putting it in (7.6) we get

$$\begin{aligned} \displaystyle \ {\left\| {u}\right\| }_{L^{t}(B_{R})}\le CR^{n/t}{\left\| {v}\right\| }_{W^{m,r}(B_{1})}\le CR^{-n\left( \frac{1}{r}-\frac{1}{t}\right) }{\left\| {u}\right\| }_{W^{m,r}(B_{R})}. \end{aligned}$$

But, because $\displaystyle t=S_{m}(r),$ we get $\displaystyle (\frac{1}{r}-\frac{1}{t})=\frac{m}{n}$ and

$$\begin{aligned} \displaystyle \ {\left\| {u}\right\| }_{L^{t}(B_{R})}\le CR^{-m}{\left\| {u}\right\| }_{W^{m,r}(B_{R})}. \end{aligned}$$

The constant C depends only on $\displaystyle n,r.$ The proof is complete. $\square$

Lemma 7.7

Let $x\in M$ and $\displaystyle B(x,R)$ be a $\epsilon$ admissible ball; we have, with $\displaystyle t=S_{m}(r),$

$$\begin{aligned} \displaystyle \forall u\in W^{m,r}(B(x,R)),\ {\left\| {u}\right\| }_{L^{t}(B(x,R))}\le CR^{-m}\ {\left\| {u}\right\| }_{W^{m,r}(B(x,R))}, \end{aligned}$$

the constant $\displaystyle C$ depending only on $\displaystyle n,\ r$, and $\displaystyle \epsilon .$

Proof

This is true in ${\mathbb {R}}^{n}$ by Lemma 7.6 so we can apply the comparison Lemma 7.5. $\square$