1

Prove that if $f:G\rightarrow \Omega$ is a one-to-one holomorphic function and $\phi:\Omega \rightarrow \mathbb{R}$ is a smooth (twice continuously differentiable) subharmonic function, then $\phi \circ f$ is subharmonic.

I have a solution that I'm not sure is correct. Rather than include every detail, I would like to propose my idea to see if it's viable. It is straightforward, using the continuity of $\phi \circ f$, $\phi$, and $f$. For $z\in G$, since $\phi$ is subharmonic at $f(z)$, we have that

$$\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z)+re^{i\theta})d\theta=\frac{1}{2\pi}\int_0^{2\pi}\Big[\phi(f(z)+re^{i\theta})-\phi(f(z))+\phi(f(z))-\phi(f(z+re^{i\theta}))+\phi(f(z+re^{i\theta}))\Big] d\theta ,$$

and now the two terms

$\phi(f(z)+re^{i\theta})-\phi(f(z))$ and

$\phi(f(z))-\phi(f(z+re^{i\theta}))$,

may be made small using continuity arguments. This leaves us with $\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z+re^{i\theta}))d\theta+\epsilon$, where $\epsilon >0$ is arbitrary, and we're done.

Note that I believe some of the hypotheses are unnecessary; this is from a qualifying exam and perhaps extra assumptions were included to make the problem seem harder. Or, perhaps, my solution is nowhere near correct :)

Thanks!

EDIT: For the sake of clarification, here is a more detailed sketch.

Let $z\in G$ be given. To show $\phi \circ f$ is subharmonic, we must find some $\gamma >0$ for which $\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z+re^{i\theta}))d\theta$ for $0<r<\gamma$. Indeed, since $\phi$ is subharmonic at $f(z)$, there exists some $\rho >0$ for which $$\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z)+re^{i\theta})d\theta$$ for $0<r<\rho$. Let $\epsilon >0$ be given.

Since $\phi$ is continuous at $f(z)$, there exists $\delta_1>0$ such that $|f(z)-y|\implies |\phi(f(z))-\phi(y)|<\frac{\epsilon}{2}$.

Since $f$ is continuous at $z$, there exists $\delta_2>0$ such that $|z-x|<\delta_2 \implies |f(z)-f(x)|<\delta_1$.

Let $\gamma=\min\{\delta_1,\delta_2\}$. Then for $0<r<\gamma$, we have that

$$|z-(z+re^{i\theta})|=r<\delta_2 \implies |f(z)-f(z+re^{i\theta})|<\delta_1 \implies |\phi(f(z))-\phi(f(z+re^{i\theta}))|<\frac{\epsilon}{2}$$

and

$$|f(z)-(f(z)+re^{i\theta})|=r<\delta_1 \implies |\phi(f(z))-\phi(f(z)+re^{i\theta})|<\frac{\epsilon}{2}.$$

Thus $$\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z)+re^{i\theta})d\theta$$

$$=\frac{1}{2\pi}\int_0^{2\pi}\Big[\phi(f(z)+re^{i\theta})-\phi(f(z))+\phi(f(z))-\phi(f(z+re^{i\theta}))+\phi(f(z+re^{i\theta}))\Big] d\theta$$

$$\leq \frac{1}{2\pi}\int_0^{2\pi}\Big[\phi(f(z+re^{i\theta}))+\epsilon \Big] d\theta$$

user122916
  • 1,227
  • 1
    For the continuity arguments, you must let $r\to 0$, don't you? But for any continuous $g$, we have $$g(w) \leqslant \frac{1}{2\pi} \int_0^{2\pi} g(w+re^{it}),dt + \epsilon$$ if we let $r$ be small enough. – Daniel Fischer Oct 02 '14 at 19:42
  • Hello, and thanks for the remark. I may be confused by your comment, so I apologize. This is not an argument for continuity, so no limits need to be involved. What needs to be shown is that, for $z\in G$, we must find some $\gamma >0$ for which $\phi(f(z))\leq \frac{1}{2\pi}\int_0^{2\pi}\phi(f(z+re^{i\theta}))d\theta$ for $0<r<\gamma$. I have edited my above post to include the full argument, but I'm not sure if it addresses your remark or is devoid of logical errors. – user122916 Oct 02 '14 at 21:15
  • 2
    Your $\gamma$ depends on $\epsilon$. That destroys the argument. For every continuous $g$, you can, for every given $\epsilon > 0$ find a $\gamma > 0$ such that for all $0 < r < \gamma$ you have $$g(w) \leqslant \frac{1}{2\pi} \int_0^{2\pi} g(w+re^{it}),dt + \epsilon.$$ The intention of the exercise is to use the characterisation of subharmonic functions in terms of the Laplacian. – Daniel Fischer Oct 02 '14 at 21:21
  • Thank you. I was able to solve it using the Laplacian. – user122916 Oct 02 '14 at 22:43

1 Answers1

1

First. I'd like to convince you that any proof that does not use the holomorphicity of $f$ is doomed. Let $\phi(x+iy) = x^2-y^2$; this is a harmonic function, hence subharmonic. Let $f(x+iy) = x+2iy$; this is an infinitely differentiable function in the real sense, but not holomorphic. The composition is: $$ \phi(f(x+iy)) = x^2-4y^2 $$ which is not subharmonic: the Laplacian is $2-8=-6$.

The key to this problem is the fact that for holomorphic $f$, the Laplacian of composition can be computed as $$\Delta (\phi\circ f) = ((\Delta \phi)\circ f )|f'|^2$$ which shows it's nonnegative. You can find the computation here.

Community
  • 1