Is there a way to show that every harmonic function is the real part of a holomorphic function without using integration equations if later theorems are allowed also?

  • 4
    $\begingroup$ This amounts to solving the Cauchy–Riemann equations as a system of PDEs. Integration is probably the only method that works generically... $\endgroup$
    – Zhen Lin
    Commented Feb 11, 2012 at 12:38

2 Answers 2


Is the following "integral free" enough?

Let $$h:\quad \Omega\to{\mathbb R}\ ,\qquad (x,y)\mapsto h(x,y)$$ be the given harmonic function which we assume to be $C^2$. Using $h$ we define the function $$f(x,y):= h_x(x,y)- i\ h_y(x,y)$$ with real part $u(x,y)=h_x(x,y)$ and imaginary part $v(x,y)=- h_y(x,y)$. As $h$ is harmonic one has $u_x\equiv v_y$, furthermore $u_y\equiv -v_x$ by equality of the mixed derivatives. So $f\colon \Omega\to{\mathbb C}$ is $C^1$ and satisfies the CR equations; therefore it is an analytic function of $z=x+iy\in\Omega$.

Assume that $\Omega$ is simply connected and chose a point $z_0\in\Omega$. Then by a standard theorem of complex analysis the function $$F(z)\ :=\ h(z_0)+\int_\gamma f(z)\ dz\ , \qquad \hbox{$\gamma\ $ a path from $z_0$ to $z$}\ ,$$ is an analytic primitive of $f$ in $\Omega$. Let $(x,y)\mapsto U(x,y)$ be the real part of $F$. Then by the CR equations, this time applied to $F$, we have $$U_x(z) -i U_y(z)=F'(z)=f(z)= h_x(z)-i h_y(z)\qquad(z\in\Omega)\ .$$ It follows that $$\nabla U(x,y)\equiv\nabla h(x,y)\qquad \bigl( (x,y)\in\Omega\bigr)\ ,$$ and as $U(x_0,y_0)=h(x_0,y_0)$ we conclude that in fact $U(x,y)\equiv h(x,y)$ in $\Omega$.

Note that we had to assume $\Omega$ simply connected. The function $h(x,y):=\log\sqrt{x^2+y^2}$ is harmonic in the punctured plane but is not the real part of an analytic function in this domain.

  • $\begingroup$ How we know that $log \sqrt{x^2+y^2}$ is not the real part of an analytic function? $\endgroup$
    – Idonknow
    Commented Oct 6, 2013 at 17:26
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    $\begingroup$ @Idonknow: This is one of the most basic facts of complex analysis. The analytic function you are looking for would be $f(z)=\log|z|+i{\rm Arg}(z)$, but this function is not well defined in all of ${\mathbb C}\setminus\{0\}$. $\endgroup$ Commented Oct 6, 2013 at 19:01
  • 1
    $\begingroup$ Salute. The clearest explanation I have seen so far. $\endgroup$
    – Dinoman
    Commented May 14, 2021 at 12:47
  • $\begingroup$ This "function" has the required real part, however, you have not shown that it is the only option. That is, one has to still show that we can't come up with some clever function that works. $\endgroup$ Commented May 20, 2023 at 19:34

The statement is false without assumptions on the domain; log$| z |$ is harmonic on the punctured plane and can locally be expressed as the real part of an analytic function, but can't be the real part of an analytic function on any annulus about $0$ since the argument can't even be continuous.

As an aside, "annulus" gets spell-checked on this site! haha


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