This is an extension of a previously asked question:

A function $f\in L^2(D)$ is weakly holomorphic if, for every $\phi\in \mathcal{C}^{\infty}_c(D)$, $$\int_D f\partial_{\bar{z}}\phi = 0.$$ I'm trying to show that each such $f$ is smooth on the interior of $D$ and is in fact a strong solution to $\partial_{\bar{z}}f=0$; i.e., $f$ is holomorphic in the usual sense.

Here's what I've proven so far:

Let $B$ be a bounded open set in $\mathbb{R}^n$ and $f\in L^p(D)$ for $1<p<\infty$. Let $g$ be a smooth, non-negative function supported in the unit ball with Lebesgue integral 1 and consider the mollifier $$f_{\epsilon}(x)=\epsilon^{-n}\int_{B}g(\frac{x-y}{\epsilon})f(y)dy$$ where $x\in B$ and $\epsilon<|x,\partial B|$. Then, $f_{\epsilon}\rightarrow f$ uniformly in the $L^p$ sense as $\epsilon\rightarrow 0$. Consequently, $f$ can be approximated by smooth, compactly supported functions in $B$.

Now, supposing this is true in the complex case if we just replace $B$ with the unit disk $D$ (I haven't proven this, but I think it's correct...) then I'm essentially done if I can demonstrate that my mollifying functions are holomorphic since I already know they're smooth. I'm not quite sure where the hypothesis would come into play, however. Anyway, is this the correct route?


  • $\begingroup$ We can always treat the unit disk $D$ in the complex plane as the two-dimensional open ball $B$ in $\mathbb R^2$: just ignore the complex structure. So, mollification works as you stated. // That said, I don't understand why you say "$f$ can be uniformly approximated". Convergence in $L^p$ norm and uniform convergence are different things. $\endgroup$ – user53153 Feb 26 '13 at 3:01
  • $\begingroup$ Yes, you're right. I think I had my wires crossed, so to speak. $\endgroup$ – Alexander Sibelius Feb 26 '13 at 3:02

It goes like this:

  1. Step back from the boundary to get an $\epsilon$ of room: $f_\epsilon$ is defined on the set $D'=\{z\in D: \operatorname{dist}(z,\partial D)>\epsilon_0\}$, and we consider $\epsilon<\epsilon_0$ below.
  2. Fubini's theorem shows that $$\int_{D'} f_\epsilon (z)\partial_{\bar z}\phi\,dz = \epsilon^{-n}\int_{|w|<\epsilon} g(w/\epsilon) \, dw \int_{D'} f(z-w)\,\partial_{\bar z}\phi(z)\,dz =0 \tag1$$ for any test function $\phi$ with compact support in $D'$.
  3. Since $f_\epsilon$ is smooth, we can integrate by parts to move the derivative over to $f_\epsilon$. Since $\int (\partial_{\bar z}f_\epsilon) \phi=0$ for all test functions, it follows that $ \partial_{\bar z}f_\epsilon=0$ on $D'$. That is, $f_\epsilon$ is holomorphic.
  4. We already knew that $f_\epsilon\to f$ in $L^p$ norm, but now this can be upgraded to uniform convergence on compact subsets of $L^p$. Indeed, for a holomorphic function $h$ we have $|h(a)|\le \frac{1}{\pi r^2}\iint_{|z-a|<r} |h(z)|$ (this can be shown in various ways, including Cauchy integral formula). Since $(f_\epsilon)$ is Cauchy in $L^p$, it is also Cauchy in the supremum norm on every compact subset of $D'$. Therefore, it converges uniformly on compact subsets of $D'$.
  5. The limit is holomorphic on $D'$, being a locally uniform limit of holomorphic functions.
  • $\begingroup$ Great, thank you very much. In my own work, I was stuck in the middle of step three. Thanks for the help! $\endgroup$ – Alexander Sibelius Feb 26 '13 at 3:29

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