Solutions of the \(\bar{\partial }\)-equation with compact support on Stein and Kähler manifold

Abstract

We study the \(\bar{\partial }\)-equation first in Stein manifold then in complete Kähler manifolds. The aim is to get \(L^{r}\) and Sobolev estimates on solutions with compact support. In the Stein case we get that for any (pq)-form \(\omega\) in \(L^{r}\) with compact support and \(\bar{\partial }\)-closed there is a \((p,q-1)\)-form u in \(W^{1,r}\) with compact support and such that \(\bar{\partial } u=\omega .\) In the case of Kähler manifold, we prove and use estimates on solutions on Poisson equation with compact support and the link with \(\bar{\partial }\) equation is done by a classical theorem stating that the Hodge Laplacian is twice the \(\bar{\partial }\) (or Kohn) Laplacian in a Kähler manifold. This uses and improves, in special cases, our result on Andreotti–Grauert-type theorem.

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Appendix

Appendix

For the reader’s convenience we shall prove certainly known results on the duality \(L^{r}-L^{r'}\) for (pq)-forms in a complex manifold.

Recall we have a pointwise scalar product and a pointwise modulus:

$$\begin{aligned} (\alpha ,\beta )\mathrm{{d}}m:=\alpha \wedge {\overline{*\beta }};\ \ \left| {\alpha }\right| ^{2}\mathrm{{d}}m:=\alpha \wedge {\overline{*\alpha .}} \end{aligned}$$

By the Cauchy-Schwarz inequality for scalar product we get:

$$\begin{aligned} \forall x\in X,\ \left| {(\alpha ,\beta )(x)}\right| \le \left| {\alpha (x)}\right| \left| {\beta (x)}\right| . \end{aligned}$$

This gives Hölder inequalities for (pq)-forms:

Lemma 6.1

(Hölder inequalities) Let \(\alpha \in L^{r}_{p,q}(\Omega )\) and \(\displaystyle \beta \in L^{r'}_{p,q}(\Omega ).\) We have

$$\begin{aligned} \left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| \le {\left\| {\alpha }\right\| }_{L^{r}(\Omega )}{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}. \end{aligned}$$

Proof

We start with \({\left\langle {\alpha ,\beta }\right\rangle }=\int \limits _{\Omega }{(\alpha ,\beta )(x)\mathrm{{d}}m(x)}\) hence

$$\begin{aligned}\ \left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| \le \int \limits _{\Omega }{\left| {(\alpha ,\beta )(x)}\right| \mathrm{{d}}m}\le \int \limits _{\Omega }{\left| {\alpha (x)}\right| \left| {\beta (x)}\right| \mathrm{{d}}m(x)}. \end{aligned}$$

By the usual Hölder inequalities for functions we get

$$\begin{aligned}&\int \limits _{\Omega }{\left| {\alpha (x)}\right| \left| {\beta (x)}\right| \mathrm{{d}}m(x)}\le {\left( {\int \limits _{\Omega }{\left| {\alpha (x)}\right| ^{r}\mathrm{{d}}m}}\right) }^{1/r}\\&\quad \times \,\left( {\int \limits _{\Omega }{\left| {\beta (x)}\right| ^{r'}\mathrm{{d}}m}}\right) ^{1/r'} \end{aligned}$$

which ends the proof of the lemma. \(\square\)

Lemma 6.2

Let \(\alpha \in L_{p,q}^{r}(\Omega )\) then

$$\begin{aligned} {\left\| {\alpha }\right\| }_{L_{p,q}^{r}(\Omega )}=\sup _{\beta \in L_{p,q}^{r'}(\Omega ),\ \beta \ne 0}\frac{\left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| }{{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}}. \end{aligned}$$

Proof

We choose \(\beta :=\alpha \left| {\alpha }\right| ^{r-2},\) then:

$$\begin{aligned} \left| {\beta }\right| ^{r'}=\left| {\alpha }\right| ^{r'(r-1)}=\left| {\alpha }\right| ^{r}\Rightarrow {\left\| {\beta }\right\| }_{L^{r'}(\Omega )}^{r'}={\left\| {\alpha }\right\| }_{L^{r}(\Omega )}^{r}. \end{aligned}$$

Hence

$$\begin{aligned} {\left\langle {\alpha ,\beta }\right\rangle }={\left\langle {\alpha ,\alpha \left| {\alpha }\right| ^{r-2}}\right\rangle }=\int \limits _{\Omega }{(\alpha ,\alpha )\left| {\alpha }\right| ^{r-2}\mathrm{{d}}m}={\left\| {\alpha }\right\| }_{L^{r}(\Omega )}^{r}. \end{aligned}$$

On the other hand we have

$$\begin{aligned} {\left\| {\beta }\right\| }_{L^{r'}(\Omega )}={\left\| {\alpha }\right\| }_{L^{r}(\Omega )}^{r/r'}={\left\| {\alpha }\right\| }_{L^{r}(\Omega )}^{r-1}, \end{aligned}$$

so

$$\begin{aligned} {\left\| {\alpha }\right\| }_{L^{r}(\Omega )}{\times }{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}={\left\| {\alpha }\right\| }_{L^{r}(\Omega )}^{r}={\left\langle {\alpha ,\beta }\right\rangle }. \end{aligned}$$

Hence

$$\begin{aligned} {\left\| {\alpha }\right\| }_{L^{r}(\Omega )}=\frac{\left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| }{{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}}. \end{aligned}$$

A fortiori for any choice of \(\beta\):

$$\begin{aligned} {\left\| {\alpha }\right\| }_{L^{r}(\Omega )}\le \sup _{\beta \in L^{r'}(\Omega )}\frac{\left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| }{{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}}. \end{aligned}$$

To prove the other direction, we use the Hölder inequalities, Lemma 6.1:

$$\begin{aligned} \forall \beta \in L_{p,q}^{r'}(\Omega ),\ \frac{\left| {{\left\langle {\alpha ,\beta }\right\rangle }}\right| }{{\left\| {\beta }\right\| }_{L^{r'}(\Omega )}}\le {\left\| {\alpha }\right\| }_{L^{r}(\Omega )}. \end{aligned}$$

The proof is complete. \(\square\)

Now we are in position to state:

Lemma 6.3

The dual space of the Banach space \(\displaystyle L_{p,q}^{r}(\Omega )\) is \(\displaystyle L_{n-p,n-q}^{r'}(\Omega ).\)

Proof

Suppose first that \(u\in L_{n-p,n-q}^{r'}(\Omega ).\) Then consider:

$$\begin{aligned} \forall \alpha \in L_{p,q}^{r}(\Omega ),\ {\mathcal {L}}(\alpha ):=\int \limits _{\Omega }{\alpha \wedge u}={\left\langle {\alpha ,{\overline{*u}}}\right\rangle }. \end{aligned}$$

This is a linear form on \(\displaystyle L_{p,q}^{r}(\Omega )\) and its norm, by definition, is

$$\begin{aligned} {\left\| {{\mathcal {L}}}\right\| }=\sup _{\alpha \in L^{r}(\Omega )}\frac{\left| {{\left\langle {\alpha ,{\overline{*u}}}\right\rangle }}\right| }{{\left\| {\alpha }\right\| }_{L^{r}(\Omega )}}. \end{aligned}$$

By use of Lemma 6.2 we get

$$\begin{aligned} {\left\| {{\mathcal {L}}}\right\| }={\left\| {{\overline{*u}}}\right\| }_{L^{r'}_{p,q}(\Omega )}={\left\| {u}\right\| }_{L^{r'}_{n-p,n-q}(\Omega )}. \end{aligned}$$

So we have \({({L_{p,q}^{r}(\Omega )})}^{*}\supset L_{n-p,n-q}^{r'}(\Omega )\) with the same norm.

Conversely take a continuous linear form \({\mathcal {L}}\) on \(\displaystyle L_{p,q}^{r}(\Omega ).\) We have, again by definition, that:

$$\begin{aligned} {\left\| {{\mathcal {L}}}\right\| }=\sup _{\alpha \in L^{r}(\Omega )}\frac{\left| {{\mathcal {L}}(\alpha )}\right| }{{\left\| {\alpha }\right\| }_{L^{r}(\Omega )}}. \end{aligned}$$

Because \({\mathcal {D}}_{p,q}(\Omega )\subset L_{p,q}^{r}(\Omega ),\) \({\mathcal {L}}\) is a continuous linear form on \(\displaystyle {\mathcal {D}}_{p,q}(\Omega ),\) hence, by definition, \({\mathcal {L}}\) can be represented by a \((n-p,n-q)\)-current u. So we have:

$$\begin{aligned} \forall \alpha \in {\mathcal {D}}_{p,q}(\Omega ),\ {\mathcal {L}}(\alpha ):=\int \limits _{\Omega }{\alpha \wedge u}={\left\langle {\alpha ,{\overline{*u}}}\right\rangle }. \end{aligned}$$

Moreover we have, by Lemma 6.2,

$$\begin{aligned} {\left\| {{\mathcal {L}}}\right\| }=\sup _{\alpha \in {\mathcal {D}}_{p,q}(\Omega )}\frac{\left| {{\left\langle {\alpha ,*\bar{u}}\right\rangle }}\right| }{{\left\| {\alpha }\right\| }_{L^{r}(\Omega )}}={\left\| {*u}\right\| }_{L^{r'}(\Omega )} \end{aligned}$$

because \(\displaystyle {\mathcal {D}}_{p,q}(\Omega )\) is dense in \(\displaystyle L_{p,q}^{r}(\Omega ).\) So we proved

$$\begin{aligned} {\left( {L_{p,q}^{r}(\Omega )}\right) }^{*}\subset L_{n-p,n-q}^{r'}(\Omega ) \hbox { with the same norm }. \end{aligned}$$

The proof is complete. \(\square\)

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Amar, E. Solutions of the \(\bar{\partial }\)-equation with compact support on Stein and Kähler manifold. Complex Anal Synerg 7, 1 (2021). https://doi.org/10.1007/s40627-021-00061-4

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Keywords

  • \(\bar{\partial }\)-equation
  • Poisson equation
  • \(L^{r}\) estimates
  • Stein
  • Riemann and Kähler manifolds