# Weighted $L^2$ Estimates for Domains $\Omega\subseteq\mathbb{C}$

EDIT: After mrf's comment below and some discussion with my instructor for the course it was decided that the below was not really an issue. Namely, I went into reading this lecture with the notion that we were going to solve the $\bar{\partial}$ equation--that this was our main goal. In other words, in the below we were mainly $f$ focused and not $\phi$ focused. In all actuality, it is the other way around. We were supposed to know that the $\bar{\partial}$ equation always has distributional solutions and that, in fact, we were really interested in finding solutions to $\bar{\partial}u=f$ with $u$ having controlled $\|\cdot\|_\phi$ norm.

This begs two questions though that I would love if someone may be able to fill in:

1. This is the one-dimensional case of Hormander's Theorem. Can someone give me intuition about why as an algebraic/differential geometer having Hormander's theorem is such a huge deal (as it is made out to be).

2. mrf says that the below theorems actually show that $\bar{\partial}u=f$ is always solvable for any $f$ since we can always find (given a fixed $f$) a $C^2(\Omega,\mathbb{R})$ subharmonic function $\phi$ for which $\displaystyle \int_\Omega\frac{|f|^2}{\Delta\phi}e^{-\phi}$ is finite (we need finiteness to actually show a solution exists). Is there an easy way to see why such a function $\phi$ always exists for a given $f$?

Thanks!

I am currently reading the Park City lecture notes on Analytic and Algebraic Geometry (this book) and am really confused by some implicit assumptions made in the first lecture of the first minicourse (Lecture 1 of Bo Berndtsson's "An Introduction to Things $\overline{\partial}$").

Let me explain some of the background to the issue I am having. Let $\phi\in C^2(\Omega,\mathbb{R})$ be subharmonic and define the inner product:

$$\langle f,g\rangle_\phi=\int_\Omega f\bar{g}e^{-\phi}$$

and the norm $\|\alpha\|_\phi^2=\langle \alpha,\alpha\rangle_\phi$. We then define $\bar{\partial}^\ast_\phi$ to be the adjoint of $\bar{\phi}$ with respect to $\langle,\rangle_\phi$. Explicitly one can show that

$$\bar{\partial}^\ast_\phi\alpha=-e^{\phi}\frac{\partial}{\partial z}\left(e^{-\phi}\alpha\right)$$

So, now we are trying to follow the proof of Theorem 1.1.3 in the book which is stated as follows:

Theorem 1.1.3 Let $\Omega\subseteq\mathbb{C}$ be a domain and suppose that $\phi\in C^2(\Omega,\mathbb{R})$ which is subharmonic. Then, for any $f\in L^2_{\text{loc}}(\Omega)$ there is a distributional solution $u$ to $\displaystyle \frac{\partial u}{\partial \bar{z}}=f$ subject to $$\int_\Omega |u|^2 e^{-\phi}\leqslant \int_\Omega \frac{|f|^2}{\Delta \phi}e^{-\phi}$$

The author states that the theorem follows from the following three propositions:

Proposition 1.1.1 Given $f$ there exists a distributional solution to $\displaystyle \frac{\partial u}{\partial\bar{z}}$ satisfying $$\|u\|_\phi^2\leqslant C\quad \mathbf{(1.3)}$$ for some $C>0$ if and only if the estimate $$\left\langle f,\alpha\right\rangle_\phi \leqslant C\|\bar{\partial}^\ast_\phi \alpha\|_\phi\quad\mathbf{(1.4)}$$ holds for every $\alpha\in C^2_c(\Omega)$.

,

Proposition 1.1.1(cont.) For any given $\mu:\Omega\to\mathbb{R}^+$ $\mathbf{(1.4)}$ holds for all $f$ satisfying $$\int_\Omega \frac{|f|^2}{\mu}e^{-\phi}\, dz\leqslant C\quad\mathbf{(1.5)}$$ if and only if $$\int_\Omega \mu|\alpha|^2 e^{-\phi}\, dz\leqslant \|\bar{\partial}^\ast_\phi\alpha\|\quad\mathbf{(1.6)}$$ holds for all $\alpha\in C^2_c(\Omega)$.

and,

Proposition 1.1.2 Let $\Omega\subseteq\mathbb{C}$ be a domain $\phi\in C^2(\Omega,\mathbb{R})$ and $\alpha\in C_c^2(\Omega)$. Then, $$\int_\Omega \Delta\phi|\alpha|^2 e^{-\phi}+\int_\Omega\left|\frac{\partial \alpha}{\partial\bar{z}}\right|^2 e^{-\phi}=\|\bar{\partial}^\ast_\phi\alpha\|\quad\mathbf{(1.7)}$$

It seems by the ease to which he claims Theorem 1.1.3 follows from these three propositions that the easy answer should be the correct one. The easier answer is that Proposition 1.1.2 shows that (1.6) holds for $\mu=\Delta\phi$. Thus, Proposition 1.1.1(cont.) implies that for every $f$ satisfying (1.5) we have that $f$ satisfies (1.4) for all $\alpha$ and thus we have a distributional solution to $\displaystyle \frac{\partial u}{\partial\bar{z}}u=f$ satisfying (1.3).

Ok, so everything seems hunky-dory, all of this goes through correctly to prove Theorem 1.1.3 if, given $f\in L^2_{\text{loc}}(\Omega)$, we could take

$$C=\int_\Omega \frac{|f|^2}{\Delta\phi}e^{-\phi}$$

The only issue is that the apply the proof of Proposition 1.1.1 we apply Riesz-Fischer to a certain operator $L$, the boundedness of which follows because we obtain a bound $\|L\|_\text{op}\leqslant C$. Thus, everything breaks down if $C$ is infinite. So, all of this strongly seems to suggest that the integral

$$\int_\Omega\frac{|f|^2}{\Delta\phi}e^{-\phi}$$

is finite for every $f\in L^2_\text{loc}(\Omega)$ and every subharmonic $\phi\in C^2(\Omega,\mathbb{R})$. But, I am fairly sure this is not true (just take $\Omega=\mathbb{C}$, $\phi=x^2+y^2$, and $f=\exp(2(x^2+y^2))$). Even if we require that $f\in L^2_{\text{loc}}(\Omega)$ and $fe^{\frac{-\phi}{2}}\in L^2(\Omega)$ (which may be a possible typo) there is still doubt that this integral always converges.

If anyone could provide any insight into what I am missing/what the author may have meant I would be extremely grateful.

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If $C$ is infinite, any distributional solution to $\frac{\partial u}{\partial \bar z} = f$ satsifies the conclusion in Theorem 1.3. – mrf Feb 17 '13 at 20:39
@mrf This makes sense, the only issue is that we don't know the existence of a distributional solution always holds! Proposition 1.1.1 doesn't just give us a solution with a bound, but the existence of a solution. – Alex Youcis Feb 17 '13 at 20:42
Ah, ok. But it's not hard to check that $E(z) = \frac1z$ is a fundamental solution to $\frac{\partial}{\partial \bar z}$ (modulo a $2\pi$ factor), so $E*f$ is a distributional solution. – mrf Feb 17 '13 at 20:45
So the point of Thm 1.3 isn't really the existence of a solution, just the bound. When you move on, you will see that it's always possible to choose $\phi$ so that that the RHS is finite (and this is the gist of the Hörmander $L^2$ solution to the $\bar\partial$-equation and the consequent solution of the Levi problem in several variables.) – mrf Feb 17 '13 at 20:48
@mrf This makes me sad. I thought this was somehow a really slick, really short proof of the existence of a solution. Is there any simple way to see that such a $\phi$ always exists? Me and a friend are giving a talk on this soon and we'd love to include that. Thanks os much!! – Alex Youcis Feb 17 '13 at 20:56

I will start with 2) In one variable this is indeed always true. To get some intution, assume that $\Omega$ is the unit disc $D$. Note that

$$\phi(z) = -\log(1-|z|^2)$$

is (strictly) subharmonic on $D$ (for example by direct computation of $\Delta\phi$). Also $\phi$ tends to $+\infty$ everywhere on $\partial D$. Furthermore, if $g(x)$ is an increasing, convex function of $x$, then $g(\phi(z))$ is also subharmonic, and by choosing $g$ wisely, we can get a subharmonic function, tending to $+\infty$ at $\partial D$ as quickly as we like, in particular quickly enough to make sure that $e^{-\phi}$ kills off any singularities of $f$ near $\partial D$. (Strict subharmonicity ensures that the $1/\Delta \phi$ factor is irrelevant.) Similar constructions van be made for any open set in $\mathbb{C}$, but this is a little deceptive.

In several complex variables, there are domains $\Omega$ (even topologically trivial ones) that don't admit any plurisubharmonic exhaustion function (i.e. psh functions tending to $+\infty$ at the boundary). The domains that do are the pseudoconvex domains, and they play an extremely important part in function theory of several complex variables.

Added A few more details. If you don't have a concrete formula for $f$, you can't expect to be able to write down a concrete formula for $\phi$. Let me return to my example, $\Omega = D$. Let $0 \le r_n \nearrow 1$ be an increasing sequence and let $\Omega_n = \{ r_n \le z < r_{n+1} \}$. Since $f \in L^2_{loc}$, the value $$M_n = \int_{\Omega_n} |f|^2$$ is finite. Choose the convex function $g$ so large and so rapidly increasing that $e^{-\phi} < 2^{-n}/M^n$ on $\Omega_n$ and so convex that $\Delta \phi$ is bounded away from $0$ on $D$. With this choice of $\phi$, the integral $$\int_D |f|^2 \frac{e^{-\phi}}{\Delta\phi} < \infty.$$

1) To be honest, $\newcommand{\dbar}{\bar\partial}L^2$-solutions of the $\dbar$-equation in one complex variable is not a big deal. In one variable, we have a rich, extremely well-developed theory and lots of methods to construct holomorphic functions with prescribed properties (Runge's theorem, Weierstrass factorizations, Mittag-Leffler's theorem, the Cauchy transform, etc. etc. etc.) In several variables the situation is very different. Let me illustrate the power of the $\bar\partial$-method with a very simple example, a (baby) version of Ohsawa-Takegoshi's extension theorem.

Theorem Assume that $\Omega \subset \mathbb{C}^n$ is pseudoconvex, and let $X = \Omega \cap (\mathbb{C}^{n-1}\times \{0\})$. If $f$ is a holomorphic function on $X$, there is an extension of $f$ to $\Omega$, i.e. a holomorphic function $F$ on $\Omega$ such that $F|_X = f$.

Note that the assumption that $\Omega$ is pseduoconvex is exactly what allows us to solve the $\dbar$-equation with estimates in a weighted $L^2$-space. For this version of Ohsawa-Takegoshi, we don't need the estimates, just the solution. Via some standard techniques, we can adapt the Hörmander solution to other function spaces. For example if $f = W^s$, then we can find a solution $u\in W^{s+1}$ (Sobolev spaces). Hence, by Sobolev's embedding theorem, if $f\in C^\infty$, there is a solution $u \in C^\infty$.

Proof Let $\Psi$ be a smooth cutoff function such that $\Psi \equiv 1$ on a (relative) open neighborhood of $X$ and $\Psi \equiv 0$ on the part of $\Omega$ that doesn't project down to $X$. Define $$F(z) = \Psi(z) f(\pi(z)) + z_nv(z)$$ where $\pi : \mathbb{C}^n \to \mathbb{C}^n$ is the natural projection onto the first $n-1$ coordinates, and $v$ is a $C^\infty$ function that remains to be chosen. Note that $F$ is well-defined on $\Omega$ and is a smooth extension of $f$ to $\Omega$. To get a holomorphic extension, we want to make sure that $\dbar F = 0$, i.e.

$$\dbar v(z) = \frac{(-\dbar\Psi(z))f(\pi(z))}{z_n}\tag{*}$$

but the right-hand side here is $C^\infty$ (since $\dbar\Psi \equiv 0$ on a neighborhood of $z_n=0$) and $\dbar$-closed. Hence we can find a smooth $v$ solving (*), and this gives us our holomorphic extension of $f$.

This particular trick: solve the problem in the $C^\infty$-category (which is often trivial) and then adjust the solution by solving a clevely chosen $\dbar$-equation is very common in SCV, and the main reason why the Hörmander machinery has turned out to be so important. In SCV, it is almost never possible to give constructions of holomorphic functions with prescribed properties, but the $\dbar$-theory often allows us to overcome this.

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Thank you very much! One question, I have been trying to make rigorous your answer to 2), but am having difficulties. In a somewhat different direction than the one you were suggesting I was able to find $\phi$ for $f$ non-vanishing holomorphic by taking the, somewhat ugly, function $\phi=\log|f|^2+\psi$ where $\psi(x,y)=x^2+y^2+x^6+y^6$ in which case it's not hard to see that $$\int_\Omega \frac{|f|^2}{\Delta\phi}e^{-\phi}\leqslant \int_\Omega \frac{1}{\Delta\psi}\leqslant \int_\mathbb{C}\frac{1}{\Delta\psi}<\infty$$ For general $f$ though I have yet to be successful. – Alex Youcis Feb 18 '13 at 20:16
@AlexYoucis Is that enough detail? – mrf Feb 19 '13 at 10:08
Yes, I think that will do. To make this rigorous, even on a general Riemann surface, do we take compact exhaustions of the space and work similarly? – Alex Youcis Feb 20 '13 at 5:24
@AlexYoucis Sorry, I forgot to answer. Yes, this should work as long as the domain, or Riemann surface, has a (pluri-)subharmonic exhaustion function. I'm not sure if every exotic Riemann surface does though. – mrf Feb 21 '13 at 18:38