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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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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