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Let $u$ be a real valued harmonic function on $\mathbb{C}$. Let $g: \mathbb{R^2} \to \mathbb{R}$ be defined by:

$$g(x,y)=\int_0^{2\pi}u(e^{i\theta}(x+iy))\sin \theta \,d\theta$$ Which of the following statements is TRUE?

(A) $$ is a harmonic polynomial

(B) $$ is a polynomial but not harmonic

(C) $$ is harmonic but not a polynomial

(D) $$ is neither harmonic nor a polynomial

Differentiating under the integral sign I see that $g$ is harmonic. But How do I conclude if it is a polynomial or not??

gebruiker
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tattwamasi amrutam
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    One thing that is true for polynomials and no other functions is that the derivative is evenually zero. I don't know what the derivative looks like, but there might be some way you can tell whether repeated differentiation will eventually kill everything or not. – Arthur Jan 30 '16 at 07:57
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    Looking at a specific example might help. For example, with $u(z)=\text{Re},z$, it seems that $g(x,y)=-\pi y$. – mickep Jan 30 '16 at 09:49

2 Answers2

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Since $u$ is real valued and harmonic on $\mathbb C,$ $u = \text { Re } f$ for some entire function $f.$ We can write $f(z)=\sum_{n=0}^{\infty}a_n z^n,$ the power series converging uniformly on compact subsets of $\mathbb C.$ Then

$$g(z) = \text { Re }\int_0^{2\pi}f(e^{i\theta}z)\sin \theta\, d \theta = \text { Re } \int_0^{2\pi}\left (\sum_{n=0}^{\infty}a_n(e^{i\theta}z)^n\right )\frac{e^{i\theta} - e^{-i\theta} }{2i}\, d \theta $$ $$=\text { Re }\frac{1}{2i}\sum_{n=0}^{\infty}a_nz^n\int_0^{2\pi}e^{in\theta}(e^{i\theta} - e^{-i\theta} )\, d \theta.$$

The interchange of integral and summation is justified by uniform convergence. The integrals in the last expression are $0$ except for $n=1,$ where the integral equals $-2\pi.$ Thus

$$g(z) = \text { Re}\left (-\frac{\pi}{i}a_1 z\right).$$

Therefore $g$ is the real part of a holomorphic polynomial of degree $1,$ and it follows that $g$ is a real harmonic polynomial of degree $1.$

zhw.
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$$g(x,y) = \int\limits_0^{2\pi}u(e^{i\theta}(x+iy))\sin\theta d\theta = \int\limits_0^{2\pi}u(e^{i\theta}(x+iy))\dfrac{e^{i\theta} - e^{-i\theta}}{2i}\dfrac {de^{i\theta}}{ie^{i\theta}}$$ $$ = -\dfrac12\oint\limits_{|z|=1}u(z(x+iy))\left(1-\dfrac1{z^2}\right)\,dz.$$ In accordance with conditions, $u(x)$ is real valued, continuous and differentiable function.
Using residue theorem, $$g(x,y) = \dfrac12\cdot2\pi i\cdot\mathop{Res}\limits_{z=0}\dfrac{u(z(x+iy))}{z^2} = \pi i\lim\limits_{z\to 0}\dfrac d{dz}u(z(x+iy)) = $$ $$\pi i(x+iy)\lim\limits_{z\to 0}u'(z(x+iy)) = \pi i(x+iy)u'(0).$$ So $g(x, y)$ is polynomial.

Yuri Negometyanov
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  • I'm not sure what you are doing. $u$ is harmonic, not holomorphic. How does the residue theorem apply to this situation? What does $u'(0)$ even mean? – zhw. Feb 05 '18 at 21:09
  • @zhw.The proof of harmonious require that. – Yuri Negometyanov Feb 06 '18 at 04:05
  • I do not understand your comment. Your proof is not correct. $u$ is given to be real and harmonic, not holomorphic. – zhw. Feb 06 '18 at 18:02