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Let $u(x,y)$ be a harmonic polynomial of real coefficients and $$f(z)=2u(\frac{z}{2}, \frac{z}{2i})-u(0,0)$$ prove that for any $z\in \mathbb{C}$, $\text{Re} f(z)=u(x,y)$.

I tried to use the condition $u_{xx}+u_{yy}=0$ to construct harmonic conjugate $v$ then deduce $f$, but it comes out a lot of integrals, and I don't know how to simplify. A similar question goes here where $u$ is degree one wrt $y$ but here it's more general. Thanks for any help.

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    You better use the fact that it $u$ is a polynomial and harmonic. There are two interesting facts that solve the problem. Every harmonic polynomial is linear combination of homogeneus harmonic polynomials. And if the polynomial is of two variables, for every degree $m$, there are only two homogeneus harmonic polynomials linearly independent, precisily the real and the imaginry part of $(x-iy)^m$. I am trying to "clean" the proof to answer the question but you can explore this right now. – 2017-02-24

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We can consider $u(x,y)$ as a sum of polynomials with all its terms of the same degree, so is, homogeneus. Consider the one of degree $m>0$, call it $u_m(x,y)$.Next, every harmonic homogeneus polynomial of two variables of degree $m$ is a linear combination of at most two linearly independent homogeneus harmonic polynomials. Such linearly independent homogeneus polynomials can be chosen to be the imaginary part and the real part of the complex polynomial $(x+iy)^m$ (Here) E.g. For degree 3 these are, $P_1(x,y)=x^3-3xy^2$ and $P_2(x,y)=3x^2y-y^3$

$$P_{m_1}(x,y)=\Re (x+iy)^m=\binom{m}{0}x^m-\binom{m}{2}x^{m-2}y^2+\binom{m}{4}x^{m-4}y^4+\,...$$

$$P_{m_2}(x,y)=\Im (x+iy)^m=\binom{m}{1}x^{m-1}y-\binom{m}{3}x^{m-3}y^3+\binom{m}{5}x^{m-5}y^5+\,...$$

Now, calculate these polynomials for $x=z/2$ and $y=z/(2i)$, with $z=\bar x+i\bar y$

$$P_{m_1}(\frac{z}{2},\frac{z}{2i})=\binom{m}{0}\frac{z^{m}}{2^{m}}-\binom{m}{2}\frac{z^{m-2}}{2^{m-2}}\frac{z^2}{(2i)^2}+\binom{m}{4}\frac{z^{m-4}}{2^{m-4}}\frac{z^4}{(2i)^4}+\,...$$

$$=\binom{m}{0}\frac{z^{m}}{2^{m}}+\binom{m}{2}\frac{z^m}{2^m}+\binom{m}{4}\frac{z^m}{2^m}+\,...=\frac{z^{m}}{2^{m}}\left(\binom{m}{0}+\binom{m}{2}+\binom{m}{4}+\,...\right)=$$

$$=\frac{z^{m}}{2^{m}}2^{m-1}=\frac{z^{m}}{2}$$

$$P_{m_2}(\frac{z}{2},\frac{z}{2i})=\binom{m}{1}\frac{z^{m-1}}{2^{m-1}}\frac{z}{(2i)}-\binom{m}{3}\frac{z^{m-3}}{2^{m-3}}\frac{z^2}{(2i)^3}+\binom{m}{5}\frac{z^{m-5}}{2^{m-5}}\frac{z^5}{(2i)^5}+\,...$$

$$=-i\binom{m}{1}\frac{z^{m}}{2^{m}}-i\binom{m}{3}\frac{z^m}{2^m}-i\binom{m}{5}\frac{z^m}{2^m}+\,...=\frac{z^{m}}{2^{m}}\left(\binom{m}{1}+\binom{m}{3}+\binom{m}{5}+\,...\right)=$$

$$=-i\frac{z^{m}}{2^{m}}2^{m-1}=-i\frac{z^{m}}{2}$$

Now,

$$u_m(x,y)=k_{m_1}P_{m_1}(x,y)+k_{m_2}P_{m_2}(x,y)$$

$$2u_m(\frac{z}{2},\frac{z}{2i})=2k_{m_1}P_{m_1}(\frac{z}{2},\frac{z}{2i})+2k_{m_2}P_{m_2}(\frac{z}{2},\frac{z}{2i})=k_{m_1}z^{m}-k_{m_2}iz^{m}$$

$$\Re\left(2u_m(\frac{z}{2},\frac{z}{2i})\right)=k_{m_1}\Re z^m+k_{m_2}\Im z^m=u_m(\bar x,\bar y)$$

Finally, $u$ is sum of homogeneus polynomials of several degrees: $u(x,y)=\sum_{m=1}^n u_m(x,y)+C$ (with $u(0,0)=C)$

$$\Re\left(f(z)\right)=\Re\left(2u(\frac{z}{2}, \frac{z}{2i})-u(0,0)\right)=\Re\left(\sum_{m=1}^n2u_m(\frac{z}{2},\frac{z}{2i})+C\right)=$$

$$=\sum_{m=1}^n\Re\left(2u_m(\frac{z}{2},\frac{z}{2i})\right)+C=\sum_{m=1}^nu_m(\bar x,\bar y)+C=u(\bar x,\bar y)$$

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    sorry for late reply, but how you get $\bar{x}, \bar{y}$? – 2017-03-02
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    $z=\bar x+i\bar y$ defined almost at the start. $x$ and $y$ are the real and imaginary part of a complex number I gave no name, but used to "extract" the two linearly independent harmonic homogeneous (real) polynomials used in the proof. Now, $\Re z^m=\Re(\bar x+i\bar y)^m$ (and the same for $\Im$) has the same form as $\Re (x+y)^m$ ($\Im$) but different variables, this is the point. – 2017-03-02
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    If you write $P_{m_1}(x,y)=\frac{1}{2}(z^m+\bar{z}^m)$ where $z=x+iy$, then it is obvious that $P_{m_1}\left(\frac{z}{2},\frac{z}{2i}\right)=\frac{z^m}{2}$, because the $\bar{z}$-term becomes zero and the $z$-term stays $z$. Similarly for $P_{m_2}$. (This is just a minor improvement. The answer is great.) – 2017-04-13
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    @user213008, Not minor :) Thank you. – 2017-04-13