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If $f:D\to D’$ is analytic and $u: D'\to R$ is harmonic then the composition of $u$ and $f$ is harmonic in $D$.

How can I show that the above statement is true/false? Can anyone help me?

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Is $f$ an analytic vector field? –  Siminore Nov 23 '12 at 15:17
    
Why should? Of $D = D' = \mathbb R^d$, then $u = \mathrm{id}$ is harmonic, but not any analytic function $f = f \circ u$ is. –  martini Nov 23 '12 at 15:21
    
@martini: The OP is looking at $u \circ f$. $u=\text{Id}$ doesn't make sense here... –  copper.hat Nov 23 '12 at 16:12
    
This question was already discussed here... –  saz Nov 23 '12 at 16:39

1 Answer 1

Locally every harmonic function $u$ is the real part of an analytic function $g$, so locally $u \circ f = \operatorname{Re} (g \circ f)$ is the real part of an analytic function, hence harmonic. A function which is everywhere locally harmonic is globally harmonic, showing that the statement is true.

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