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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?

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    $\begingroup$ 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 Feb 11 '12 at 12:38
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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.

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  • $\begingroup$ How we know that $log \sqrt{x^2+y^2}$ is not the real part of an analytic function? $\endgroup$ – Idonknow Oct 6 '13 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 ${\mathbb C}\setminus\{0\}$. $\endgroup$ – Christian Blatter Oct 6 '13 at 19:01
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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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