# Composition of a harmonic function with a holomorphic function is still harmonic

If $$f$$ is a harmonic function in a domain $$D \subset \mathbb{C}$$, and $$g$$ is a conformal mapping of a domain $$D_0$$ onto $$D$$, is $$f \circ g$$ harmonic in $$D_0$$?

I noticed this question while reading several pdf of lecture notes, and I'm having trouble figuring it out. Can anyone help?

Thank you so much!

The answer is yes, and we only need $g$ to be holomorphic. One can prove this by directly computing the Laplacian of $f\circ g$ using the Chain Rule. I'd rather use $z$ and $\bar z$ than $x$ and $y$ for this purpose.

$$\Delta(f\circ g)=\frac{1}{4}(f\circ g)_{z\bar z} = \frac{1}{4}[(f_z\circ g) g']_{\bar z} = \frac{1}{4}(f_{z\bar z}\circ g) \overline{g'} g'= [(\Delta f)\circ g]|g'|^2=0$$

• One has to be more careful with the computation of the derivative since $f$ generally is a function of $z,\overline{z}$, so the chain rule should give you more. The result is correct though. Apr 15, 2014 at 1:18
• @user31373 The second derivative step should produce one more term f_z(g)g_{z\overline{z}}$.It seems that this term is not always zero.Moreover$\Delta u=4_{z\head{z}}$May 8, 2014 at 14:36 Let $$\phi(u,v)$$ be harmonic in $$D$$. Let $$w=f(z)=u(x,y)+iv(x,y)$$ be analytic in $$D_0$$ defining a mapping $$D_0\to D$$. We have $$\phi_x=\phi_u u_x+\phi_v v_x$$ $$\phi_y=\phi_u u_y+\phi_v v_y$$ $$\phi_{xx}=\phi_{uu}(u_x)^2+\phi_{uv} u_x v_x +\phi_u u_{xx} +\phi_{vv} (v_x)^2+\phi_{vu} v_x u_x +\phi_v v_{xx}$$ $$\phi_{yy}=\phi_{uu}(u_y)^2+\phi_{uv} u_y v_y +\phi_u u_{yy} +\phi_{vv} (v_y)^2+\phi_{vu} v_y u_y +\phi_v v_{yy}$$ $$\phi_{xx}+\phi_{yy}=[(u_x)^2+(v_x)^2][\phi_{uu}+\phi_{vv}]$$ because $$u_{xx}+v_{yy}=0$$, $$v_{xx}+v_{yy}=0$$, $$u_xv_x=-u_yv_y$$. Hence $$\phi_{uu}+\phi_{vv}=0$$ implies $$\phi_{xx}+\phi_{yy}=0$$. We conclude that $$\phi(x,y)$$ is a harmonic funtion. • This is not quite what was asked: you have$\phi = \Phi \circ f$and conclude that if$\phi$is harmonic and$f$is analytic then$\Phi$is harmonic. The question went the other way: if$\phi$is harmonic and$g$is analytic then$\phi \circ g$is harmonic. May 31, 2012 at 7:11 • @RobertIsrael In the original question,$g$is assumed to be conformal (which I read as biholomorphic), so Valentin's answer is sufficient to answer that. Of course, the assumption of invertibility is extraneous. – user31373 May 31, 2012 at 15:09 The term "harmonic" will mean "real and harmonic" below. I will use the following: 1) Compositions of holomorphic functions are holomorphic. 2) The real and imaginary parts of a holomorphic function are harmonic. 3) If$D\subset \mathbb C$is an open disc, and$u$is harmonic in$D,$then there exists$v$harmonic in$D$such that$u+iv$is holomorphic in$D.$1) is just the chain rule. 2) follows from the Cauchy-Riemann equations. If you are unfamiliar with 3), I can add a proof later. I'll assume it for now. Thm: Suppose$\Omega_1, \Omega_2 \subset \mathbb C$are open. Assume$u$is harmonic on$\Omega_2$and$f: \Omega_1 \to \Omega_2$is holomorphic. Then$u\circ f$is harmonic on$\Omega_1.$Proof: Let$a \in \Omega_1.$Because harmonicity is a local property, it suffices to show$u\circ f$is harmonic in a neighborhood of$a.$Choose an open disc$D$centered at$f(a)$contained in$\Omega_2.$Then by 3) above, there exists a$v$harmonic in$D$such that$u+iv$is holomorphic in$D.$Let$\omega = f^{-1}(D).$Then$\omega $is an open neighborhood of$a$contained in$\Omega_1,$and$f:\omega \to D.$By 1),$(u+iv)\circ f$is holomorphic in$\omega.$By 2), the real part of$(u+iv)\circ f$is harmonic in$\omega.$But this real part is precisely$u\circ f,$and we're done. Your question can be interpreted in the greater context of "maps preserving harmonic functions". Definition Let$(M,g)$and$(N,h)$be Riemannian manifolds. A mapping$\Phi:M\to N$is said to be a harmonic morphism if whenever$u:N\to\mathbb{R}$is a harmonic function (solving$\triangle_h u = 0$where$\triangle_h$is the Laplace-Beltrami operator for the Riemannian metric$h$) the composition$u\circ \Phi$is a harmonic function on$M$. Theorem A mapping is a harmonic morphism if and only if it is a harmonic map which is weakly horizontally conformal. (Don't worry too much about the undefined terms in the above theorem.) Corollary If$M$and$N$have the same number of dimensions, then • If dimension is 2,$\Phi$is a harmonic morphism if and only if$\Phi$is conformal. • If the dimension is bigger than 2,$\Phi$is a harmonic morphism if and only if$\Phi\$ is a conformal map with a constant coefficient of conformality.

For reference, see this paper of Bent Fuglede's.

• A typo in the Theorem creates ambiguity: should "with is" be read as "if it is" or "if and only if it is"?
– user31373
May 31, 2012 at 15:11
• @Leonid: fixed. May 31, 2012 at 15:42

I was wondering the same question and reached here.

Here is how I thought about it:

Let $$u$$ be harmonic. And let $$v$$ be its harmonic conjugate. Then $$f(z) = u(z) + iv(z)$$ is analytic. If $$g$$ is another analytic function such that range of $$g$$ is contained in the domain of $$f$$ then $$f(g(z)) = u(f(z)) + iv(g(z))$$ is also analytic with $$Re (f(g(z))) = u(f(z))$$.