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May i ask you for a little help about a problem with harmonic function? It seems to be not that difficult, in a way even intiutively obvious but i don't really know how to show this explicitly.

We have $U\subset \mathbb{R}^{2}$ open, non-empty, connected subset and a harmonic function $u: U\rightarrow \mathbb{R}$ given by $f:=\frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y}$. Let $F$ be a holomorphic function $U\rightarrow \mathbb{C}$ with $F'=f$. Show that there exist an $c\in \mathbb{R}$ such that $\operatorname{Re}(F) = u + c$.

Okay, $f$ is given only through $u$. This means that after the integration of $F$ i will have an expression with the real part of $u(x,y)$.

I would be glad if someone could help me to move on. Thank you in advance!

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I don't even think this is true as stated. I feel as though we may need some connectivity conditions on $U$. If we assume that $U$ is simply connected this is not bad, for example. – Alex Youcis Mar 9 '13 at 23:29
@AlexYoucis, thank you for the correction! $U$ must be open, nonempty and connected. I edited my first post. – Lullaby Mar 9 '13 at 23:32
I have posted an answer! – Alex Youcis Mar 9 '13 at 23:34
Just connected is not enough. You need that $U$ is simply connected, otherwise $F$ might not exist. – mrf Mar 9 '13 at 23:36
@mrf There is no issue if you are given the $f$. But yes, we clearly need simply connected if we want to guarnatee the existence of such an $f$. – Alex Youcis Mar 12 '13 at 16:24

Your formulation is a little strange. I assume that you start with a harmonic function $u$ and then define $f$.

First note that $f$ is holomorphic since it satisfies Cauchy-Riemann's equations. Assuming that $U$ is simply connected (and this assumption is essential), $f$ has a primitive, $F$. Again, using Cauchy-Riemann, it follows that if $F=a+ib$, you have that $a'_x = \operatorname{Re} f = u'_x$ and similarly $a'_y = u'_y$, so using that $U$ is connected, $a = u+c$.

(This is one common way to prove the existence of harmonic conjugates.)

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Hello mrf, I have the same problem as the OP, but I don't understand your answer, specially when you say "using that $U$ is connected, $a = u + c$. How could you use this assumption to conclude $a= u+c$. Sorry if my question sounds stupid, but I'm not used to work with connectedness of sets. – Bman72 Mar 8 '14 at 10:31
I've also another question, why can you say that f has a primitive because $U$ is connected, does it follow from some definition? – Bman72 Mar 8 '14 at 10:35
Hi @mrf, if I have a harmonic function with known boundary values, it apparently must be unique. I am thinking that this is true, just from looking at the Poisson integral formula, which gives its values on the interior, from knowing its values on the boundary. However, why is the Maximum Principle needed here to argue the uniqueness of the harmonic function? I've seen it show up on several answers, but I don't see why it's necessary. Thanks professor :-) – User001 Dec 17 '15 at 19:05

Hint: Apply the Cauchy-Riemann equations to conclude that the total derivative of $\text{Re}(F)$ is equal to the total derivative of $u$. Use then connectivity and the Intermediate Value Theorem to conclude that $u$ and $\text{Re}(F)$ differ by a constant.

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