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Let $f=(u,v)\in \mathscr{D}'(U,\mathbb{C})$ be a distribution, where $U\subset\mathbb{C}=\mathbb{R}^2$ is an open set and $u$ and $v$ are the projection of $f$ onto the real and imaginary axis (ie $\langle f,\phi\rangle=\langle u,\phi\rangle+i\langle v,\phi\rangle$). Suppose that $$ \frac{\partial}{\partial \overline{z}}f=0\qquad\text{in U,} $$ where $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\bigg(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\bigg)$ and the derivatives are in distributional sense. Does it follow that $f$ is holomorphic in the classical sense, ie $f\in C^\infty(U,\mathbb{C})$ and the Cauchy-Riemann equations are satisfied?

The obvious idea would be to mollify, get holomorphic functions and then take the limit. But how can we conclude that the limit is still holomorphic?

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    $\begingroup$ That is called the Weyl lemma. $\endgroup$
    – Ben McKay
    Commented Feb 1, 2022 at 16:45
  • $\begingroup$ @BenMcKay Doesn't the Weyl Lemma require $f\in L^1_{loc}$? $\endgroup$
    – No-one
    Commented Feb 5, 2022 at 11:39
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    $\begingroup$ no: see Folland's Real Analysis, p. 308. $\endgroup$
    – Ben McKay
    Commented Feb 5, 2022 at 11:49
  • $\begingroup$ Thanks. I knew the proof on Folland for tempered distributions but I never noticed that the same proof also holds for distributions in $\mathscr{D}'$. Very nice! $\endgroup$
    – No-one
    Commented Feb 5, 2022 at 12:14

2 Answers 2

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The trick is to replace the Cauchy integral formula $f_n(z)=\frac1{2i\pi}\int_{|s-a|=r} \frac{f_n(s)}{s-z}ds$ on a circle

by one on an annulus $$f_n(z)=\int_r^R \psi_{r,R}(t)\frac1{2i\pi}\int_{|s-a|=t} \frac{f_n(s)}{s-z}ds dt\label{1}\tag{1}$$

where $f_n= f\ast \phi_n$ is a mollified version of $f$, so $f_n$ is smooth and holomorphic on a slightly smaller open $U_n$, and $\psi_{r,R}\in C^\infty_c(r,R),\int \psi_{r,R}=1$.

Equation \eqref{1} is an integral of $f_n$ against a $C^\infty_c(\Bbb{C})$ function $h_z$ where (locally) $z\to h_z$ is continuous in the $C^\infty_c$ topology,
so there won't be any problem when letting $n\to \infty$, obtaining the local uniform convergence of $f_n$ to an analytic function representing $f$.

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  • $\begingroup$ To obtain the local uniform convergence at last line you're using that $f_n\to f$ with respect to the strong topology of $\mathscr{D}'$, right? $\endgroup$
    – No-one
    Commented Feb 3, 2022 at 19:38
  • $\begingroup$ (for $|z-a|<r$ with $r$ small enough so that the disk is contained in some $U_n$) $z\to \langle f,h_z\rangle$ is a continuous and clearly holomorphic function and $f_n(z)= \langle f, \phi_n\ast h_z\rangle$ converges to it pointwise and in the sense of disttributions. @Titti $\endgroup$
    – reuns
    Commented Feb 3, 2022 at 19:44
  • $\begingroup$ I agree, but we want to show $f=[z\mapsto \langle f,h_z\rangle]$ as a distribution. For a test function $\varphi$, we know $\langle\langle f_n(w),h(z,w)\rangle,\varphi(z)\rangle=\langle f_n(z),\varphi(z)\rangle\to\langle f(z),\varphi(z)\rangle$. But to conclude we must also show $\langle\langle f_n(w),h(z,w)\rangle,\varphi(z)\rangle\to\langle\langle f(w),h(z,w)\rangle,\varphi(z)\rangle$, and this is true because $\langle f_n(w),h(z,w)\rangle\to \langle f(w),h(z,w)\rangle$ uniformly in $z$ (since $f_n\to f$ wrt the strong topology of $\mathscr{D}'$). $\endgroup$
    – No-one
    Commented Feb 3, 2022 at 20:26
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I've just realized that, if $f$ is in $L^1_{loc}$ and not just in $\mathscr{D}'$, my question can be answered using Weyl's lemma for harmonic functions. Indeed, from $\frac{\partial f}{\partial \overline{z}}=0$ it easily follows that $\Delta u=\Delta v=0$, and then Weyl's lemma implies that $u$ and $v$ are smooth. But then $f$ is smooth and satisfies the Cauchy-Riemann equations, thus is holomorphic.

EDIT: As pointed out by Ben McKay in the comments, the hypothesis that $f\in L^1_{loc}$ is not necessary.

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