Exact sequences and estimates for the $\overline{\partial }$-problem

Abstract

We study Sobolev estimates for solutions of the inhomogenous Cauchy–Riemann equations on annuli in ${\mathbb {C}}^n$, by constructing exact sequences relating the Dolbeault cohomology of the annulus with respect to Sobolev spaces of forms with those of the envelope and the hole. We also obtain solutions with prescribed support and estimates in Sobolev spaces using our method.

Appendix: Non-Hausdorff functional analysis

In this appendix we collect together definitions and results about the non-Hausdorff topological vector spaces that arise in the study of cohomology groups. Much of this material consists of routine variations on well-known results. Proofs are included only when they are sufficiently different from the classical situation.

1.1 Semi-inner-product spaces

1.1.1 Definitions and basic properties

By a semi-normed (linear) space, we mean a complex vector space V along with a distinguished seminorm $\left\| \cdot \right\| :V\rightarrow {\mathbb {R}}$ (see [37, Definition 7.3, page 59]). The semi-normed space $(V, \left\| \cdot \right\| )$ a becomes a (not necessarily Hausdorff) topological vector space under the natural semi-metric $d(x,y)= \left\| x-y\right\|$. Using the semi-metric, one defines a topology on V: a basis for the topology consists of the semi-balls

$$\begin{aligned} B(x,\epsilon )=\{y\in X\mid d(x,y)<\epsilon \}.\end{aligned}$$

However, the topology so obtained is not necessarily Hausdorff. For example, the closure of the singleton $\{x\}$ consists of all $y\in V$ such that $d(x,y)=0$, which may be true for a point different from x.

The semi-normed space $(V, \left\| \cdot \right\| )$ is a semi-inner-product space (SIP space for short) if there is a semi-inner-product $(\cdot , \cdot )$ on V such that $\left\| x\right\| ^2=(x, x)$ for each $x\in V$, where a semi-inner product is a sesquilinear, hermitian-symmetric form on V which satisfies $(x,x)\ge 0$. (It is not assumed that $(x,x)=0$ only if $x=0$). It is not difficult to check that the Cauchy–Schwarz inequality and the triangle inequality continue to hold in a SIP space (though the conditions under which equality holds in these inequalities need to be modified). By the Cauchy–Schwarz inequality, we have the following: If X is a SIP space and $z\in X$ is such that $\left\| z\right\| =0$, then for all $x\in X$, we have $\left\langle x,z \right\rangle =0.$

We say that a semi-normed space V is complete if each Cauchy sequence converges, i.e., if $\{w_j\}\subset V$ is such that $\lim _{j,k\rightarrow \infty } \left\| w_j-w_k\right\| =0$ then there is $w\in V$ such that $w_j\rightarrow w$. A SIP space which is complete in its semi-norm is said to be a semi-Hilbert space.

The following elementary result characterizing continuous maps is proved in the same way as the Hausdorff case:

Proposition A.1

Let $T:X\rightarrow Y$ be a linear map of SIP spaces. Then T is continuous if and only if there is a $C\ge 0$ such that $\left\| Tx\right\| _Y \le C \left\| x\right\| _X$.

Given two SIP spaces X and Y, there is a natural semi-inner product on the algebraic direct sum $X\oplus Y$, given by

$$\begin{aligned} \left\langle (x,y),(x',y') \right\rangle _{X\oplus Y}= \left\langle x,x' \right\rangle _X + \left\langle y,y' \right\rangle _Y. \end{aligned}$$

(A.1)

Notice, however, that if $Z=X\oplus Y$ is a direct sum of SIP spaces, then we do not have in general that $Y=X^\perp =\{z\in Z : \left\langle z,x \right\rangle =0 \text { for all } x\in X\}$. Indeed if X is indiscrete, then $X^\perp =Z$.

The following proposition, versions of which are widely known, describes the structure of a general SIP space as the sum of an indiscrete and a Hausdorff part. The indiscrete part ${\textsf {Ind}}(X)$ and the reduced form ${\textsf {Red}}(X)$ of a semi-inner-product space are defined as in (1.3) and (1.4). We establish the linear homeomorphism (1.2):

Proposition A.2

Let $(X, \left\langle \cdot ,\cdot \right\rangle )$ be a SIP space. Then ${\textsf {Ind}}(X)$ is the unique indiscrete closed linear subspace of X such that if Z is any linear subspace of X algebraically complementary to ${\textsf {Ind}}(X)$ then

• (1) restricted to Z, the semi-inner product of X is an inner product, so Z is Hausdorff.

• (2) the semi-inner product space ${\textsf {Ind}}(X)\oplus Z$, thought of as a topological vector space, is linearly homeomorphic to X.

Proof

Recall that ${\textsf {Ind}}(X) =\{x\in X: \left\| x\right\| =0\}$ by (1.3). Then ${\textsf {Ind}}(X)$ is a closed linear subspace of X and with the restricted semi-norm, ${\textsf {Ind}}(X)$ is indiscrete. Let Z be a linear subspace of X which is algebraically complementary to ${\textsf {Ind}}X$, i.e., each element of X can be uniquely written as $y+z$ where $y\in {\textsf {Ind}}(X)$ and $z\in Z$.

If $z\in Z$ is such that $\left\| z\right\| =0$, then $z\in {\textsf {Ind}}(X)$. But since ${\textsf {Ind}}(X)\cap Z=\{0\}$ it follows that $z=0$. Consequently, Z is an inner-product space.

Let the map $P:X\rightarrow {\textsf {Ind}}(X)$ be defined by $Px=y$, where $x=y+z$ is the unique representation of x as the sum of an element $y\in {\textsf {Ind}}(X)$ and $z\in Z$, so that P is the projection map onto ${\textsf {Ind}}(X)$ and has kernel Z. Then P is continuous, since ${\textsf {Ind}}X$ has the indiscrete topology. The map

$$\begin{aligned} X \rightarrow {\textsf {Ind}}(X) \oplus Z\end{aligned}$$

given by

$$\begin{aligned} x \mapsto (Px, x-Px)\end{aligned}$$

is clearly injective, and it is continuous with respect to the natural topology on ${\textsf {Ind}}(X) \oplus Z$, since each of the two components is continuous. The map ${\textsf {Ind}}(X)\oplus Z \rightarrow X$ given by $(y,z)\mapsto y+z$ is its continuous linear inverse, so this is a linear homeomorphism. The result follows. $\square$

1.1.2 Quotients

Let X be a TVS (topological vector space) and $Z\subset X$ a linear subspace of X. The pair $(X/Z, \pi )$ where X/Z is the quotient topological vector space and $\pi :X\rightarrow X/Z$ is the continuous natural projection enjoys the following universal property: if $f:X\rightarrow W$ is a continuous linear map of topological vector spaces such that $f|_Z=0$, then there is a unique continuous linear map $\overline{f}: X/Z\rightarrow W$ such that $f=\overline{f}\circ \pi$, i.e. the following diagram commutes:

Notice that it follows from this property that as a linear space, X/Z can be identified with the quotient vector space, defined in the usual way (see [37, p. 15]). The following proposition shows that SIP spaces, unlike inner-product spaces, are stable with respect to quotients.

Proposition A.3

Let $(X,(\cdot ,\cdot ))$ be an SIP space, and let $Z\subset X$ be a linear subspace. There is a semi-inner product $\langle \cdot , \cdot \rangle$ on the quotient vector space X/Z such that the semi-norm topology on $(X/Z, \langle \cdot ,\cdot \rangle )$ coincides with the quotient topology.

Proof

Let $\pi :X\rightarrow X/Z$ be the quotient map. We define

$$\begin{aligned} \left\| \pi (x)\right\| _{X/Z}= \inf _{z\in Z} \left\| x+z\right\| , \end{aligned}$$

(A.3)

which is defined on all of X/Z since $\pi$ is surjective, and is easily seen to be a seminorm using standard arguments. Let $f:X\rightarrow W$ be a continuous linear map of topological vector spaces such that $f|_Z=0$, and let $\overline{f}:X/Z\rightarrow W$ be the induced map. Now notice that for each $x\in X$ and $z\in Z$ we have for a $C\ge 0$ depending only on the map f:

$$\begin{aligned}\left\| \overline{f}(\pi (x))\right\| _W= \left\| f(x)\right\| _W = \left\| f(x+z)\right\| _W \le C \left\| x+z\right\| _X.\end{aligned}$$

Since this holds for each $z\in Z$, it follows that

$$\begin{aligned} \left\| \overline{f}(\pi (x))\right\| _W \le C \inf _{z\in Z}\left\| x+z\right\| _X= C\left\| \pi (x)\right\| _{X/Z},\end{aligned}$$

which shows that $\overline{f}$ is continuous, and therefore by the universal property (A.2), the seminorm on X/Z generates the quotient topology. To complete the proof we only need to show that this semi-norm is generated by a semi-inner-product. A classic argument well-known for norms and inner products (see [23]) shows that a semi-norm $\left\| \cdot \right\| _V$ on a vector space V is generated by a semi-inner product if and only if for all $x,y\in V$ we have the parallelogram identity:

$$\begin{aligned} \left\| x+y\right\| ^2_V + \left\| x-y\right\| ^2_V = 2 \left\| x\right\| ^2_V + 2 \left\| y\right\| ^2_V. \end{aligned}$$

(A.4)

Now, let $x,y\in X$ and $z,w\in Z$, and apply (A.4) to $x+z$ and $y+w$ in X. Then

$$\begin{aligned} \left\| x+y+(z+w)\right\| ^2 +\left\| x-y+(z-w)\right\| ^2 = 2 \left\| x+z\right\| ^2 + 2 \left\| y+w\right\| ^2. \end{aligned}$$

Let $u=z+w$ and $v=z-w$. Then, as z and w range independently over Z, so do u and v. Therefore, we conclude by taking infima of both sides of the above equation that

$$\begin{aligned} \left\| \pi (x)+\pi (y)\right\| _{X/Z}^2 + \left\| \pi (x)-\pi (y)\right\| _{X/Z}^2 = 2\left\| \pi (x)\right\| _{X/Z}^2 +2\left\| \pi (y)\right\| _{X/Z}^2.\end{aligned}$$

Therefore, since the parallelogram identity holds for the seminorm, the quotient X/Z is a semi-inner-product space. $\square$

The following fact is proved in the same way as in the classical situation (cf. [37, Proposition 4.5, page 34]):

Proposition A.4

Let X be a SIP space and Z a linear subspace. Then the semi-inner-product on X/Z is an inner-product (equivalently, X/Z is Hausdorff) if and only if Z is closed in X.

In view of Proposition A.2, for a semi-inner-product space X, the indiscrete part of X is easily verified to be a closed subspace of X, and each closed subspace of X contains the subspace ${\textsf {Ind}}(X)$. The reduced form of X defined by (1.4) is a normed (and therefore Hausdorff) space, thanks to Proposition A.4.

1.2 Short exact sequences of semi-inner-product spaces

Let

$$\begin{aligned} 0\rightarrow X\xrightarrow {\rho }Y \xrightarrow {\lambda }Z\rightarrow 0 \end{aligned}$$

(A.5)

be a short exact sequence of semi-inner-product spaces and continuous maps. We collect here a few simple observations about such sequences:

• (1) As with any exact sequence of vector spaces, we have an algebraic isomorphism of vector spaces $Y/\rho (X)\cong Z$, and an algebraic splitting, i.e., a direct sum representation of vector spaces

  $$\begin{aligned} Y = \rho (X) \oplus \mu (Z), \end{aligned}$$

  (A.6)

  where $\mu :Z\rightarrow Y$ is an injective linear map. Notice that the splitting is not natural, i.e., the map $\mu$ is not determined by the exact sequence (A.5). We emphasize that this splitting is not topological, i.e, the topology on Y may be different from the direct sum topology from the subspaces $\rho (X)$ and $\mu (Z)$.

• (2) Since $\lambda :Y\rightarrow Z$ is continuous and vanishes on $\rho (X)$ by the universal property of quotient TVS (see Sect. A.1.2 above), we obtain an induced continuous linear bijection

  $$\begin{aligned} \overline{\lambda }: (Y/\rho (X))^{\mathrm {top}} \rightarrow Z, \end{aligned}$$

  (A.7)

  where $(Y/\rho (X))^{\mathrm {top}}$ is the vector space $Y/\rho (X)$ endowed with the quotient topology.

• (3) Assume now that in (A.5) the space Z is Hausdorff.

  Then since $\overline{\lambda }$ is continuous and injective, it follows that $(Y/\rho (X))^{\mathrm {top}}$ is Hausdorff, and by the closed-graph theorem, $\overline{\lambda }$ is an isomorphism.

  Since $(Y/\rho (X))^{\mathrm {top}}$ is Hausdorff, Proposition A.4 implies that $\rho (X)$ is closed in Y, and therefore must contain the indiscrete subspace ${\textsf {Ind}}(Y)$ of Y (see (1.3)). Thanks to Proposition A.2, we have a direct sum decomposition of TVS

  $$\begin{aligned}\rho (X)= {\textsf {Ind}}( Y)\oplus V, \end{aligned}$$

  where V is a linear subspace of $\rho (X)$ algebraically complementary to ${\textsf {Ind}}(Y)$, i.e., each $x\in \rho (X)$ has a unique representation $x=y+z$, where $y\in {\textsf {Ind}}( Y)$ and $z\in V$. Let W be an algebraic complement of ${\textsf {Ind}}( Y)$ in Y such that $V\subset W$ (such a complement exists for algebraic reasons). Then clearly W is a Hilbert space in the inner product induced from Y and V is a closed subspace of W, so we have an orthogonal decomposition $W=V\oplus V'$, where $V'$ is the orthogonal complement of V in W. Then we have a direct sum decomposition of TVS

  $$\begin{aligned} Y= \rho (X) \oplus V'.\end{aligned}$$

  It follows by exactness of (A.5) that the restriction $\lambda |_{V'}:V'\rightarrow Z$ is a bijective continuous linear map of Hilbert spaces, and therefore an isomorphism in the category of TVS by the closed-graph theorem. If we now let $\mu :Z\rightarrow Y$ be the inverse of $\lambda |_{V'}$, we again obtain a splitting as in (A.6), but (a) now the splitting is topological, i.e., the topology on Y is the same as the direct sum topology from the right hand side, (b) though the splitting is still not natural, since the algebraic complement W cannot be chosen naturally.

• (4) Assume now that in (A.5) both the spaces Y and Z are Hausdorff.

  Since $\rho$ is continuous and injective and Y is Hausdorff, it follows that X is also Hausdorff. Therefore, we have, by standard results in the theory of Hilbert spaces, an orthogonal direct sum representation

  $$\begin{aligned} Y= {\text {img}}\rho \oplus ({\text {img}}\rho )^\perp = {\text {img}}\rho \oplus (\ker \lambda )^\perp = {\text {img}}\rho \oplus {\text {img}}(\lambda ^\dagger ),\end{aligned}$$

  where $\lambda ^\dagger :Z\rightarrow Y$ is the Hilbert-space adjoint of $\lambda$. Therefore we have a natural orthogonal splitting

  $$\begin{aligned} Y= \rho (X)\oplus \lambda ^\dagger (Z). \end{aligned}$$

  (A.8)

1.3 Cochain complexes

In this paper we consider cochain complexes of the form

$$\begin{aligned} \rightarrow E^{q-1}\xrightarrow {d^{q-1}} E^q \xrightarrow {d^q} E^{q+1} \xrightarrow {d^{q+1}}\end{aligned}$$

where each $E^q$ is an inner-product space and the differentials $d^q$ are continuous linear maps satisfying $d^q\circ d^{q-1}=0$ for each q. Note that it is not assumed that the space $E^q$ is complete in the norm induced by the inner product. It is of course possible to consider much more general topologized cochain complexes (see [27]). The cohomology groups of the complex are the quotient vector spaces

$$\begin{aligned} H^{q}(E) = \frac{Z^{q}(E)}{B^q(E)},\end{aligned}$$

(A.9)

where $Z^{q}(E)= \ker d^q$ and $B^q(E)= {\text {img}}d^{q-1}$ are the spaces of cocycles and coboundaries respectively. Since $d^q$ is continuous, $Z^q(E)$ is a closed subspace of $E^q$. The following is clear:

Proposition A.5

If E is a cochain complex of inner-product spaces, then the cohomology group $H^q(E)$ of (A.9) has a natural structure of a semi-inner product space, and this semi-inner-product gives rise to the quotient topology. This semi-inner product space is an inner product (and therefore Hausdorff) space if and only if $B^q(E)$ is closed as a subspace of $Z^q(E)$. Further, the inner product space $H^q(E)$ is a Hilbert space provided that the space $E^q$ is a Hilbert space.

Let (E, d) and $(F,\delta )$ be cochain complexes of inner-product spaces. A continuous cochain map f is given by continuous linear maps $f^q: E^q\rightarrow F^q$ such that for each q we have $\delta ^{q}\circ f^q = f^{q+1}\circ d^q$, i.e. the following diagram commutes for each q:

It is well-known (see [25, Chapter XX]) that such a map induces a linear map of the cohomologies in each degree. The functoriality of cohomology interacts nicely with the topology:

Proposition A.6

Let $f: (E,d)\rightarrow (F,\delta )$ be a continuous cochain map between cochain complexes of inner-product spaces. Then the induced map $f^q_*:H^q(E)\rightarrow H^q(F)$ is continuous for each q.

Proof

Consider the following diagram, where $\pi _E$ and $\pi _F$ denote the natural continuous projections onto the cohomology groups, and which commutes by the definition of the induced map $f^q_*$:

Using the universal property of quotients as in diagram (A.2), since $\pi _F\circ f^q$ is a continuous map from $Z^q(E)$ to $H^q(F)$ which vanishes on $B^q(E)$ it follows that the induced map $f^q_*$ is continuous. $\square$