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On pp. 81-82, I have two questions:
- Why do the Cauchy-Riemann equations imply that all complex derivatives except the $m^{th}$ vanish at the origin? If anything, this seems to be a consequence of homogeneity: you can only have terms of $m^{th}$ degree if there's a $z^m$ in the complex function for which the polynomial is the real part.
- Why does "this" imply that $H_m(\mathbb{R}^n)$ is the complex linear span of $\{z^m, \overline{z}^m\}$? The author shows that $p = cz^m + \overline{cz^m}$ for some constant $c$, so that we would seem not to be free to take arbitrary linear combinations of $\{z^m, \overline{z}^m\}$ -- only combinations in which one coefficient is the complex conjugate of the other.
Let me write $f=p+ig$. Since $f$ is holomorphic, you automatically obtain that $\partial_{\overline{z}}f \equiv 0$ (recall that this condition is equivalent to the fact that $f$ satisfies the Cauchy-Riemann equations). Now we deal with $\partial_{z}f$. The Cauchy-Riemann equations says $$\partial_{x}p=\partial_{y}g \quad \partial_{y}p=-\partial_{x}g.$$ Thus \begin{align*} \partial_{z}f&=\frac{1}{2}\left( \partial_{x}f+\frac{1}{i}\partial_{y}f \right) \\ &=\frac{1}{2}\left( \partial_{x}p+i\partial_{x}g+\frac{1}{i}\partial_{y}p+\partial_{y}g \right) \\ &=\frac{1}{2}\left( \partial_{x}p-i\partial_{y}p+\frac{1}{i}\partial_{y}p+\partial_{x}p \right) \\ &=\partial_{x}p+\frac{1}{i}\partial_{y}p \end{align*} This says that $\partial_{z}f$ is a homogeneous polynomial of degree $m-1$ (because the derivative of a homogeneous polynomial is also a homogeneous polynomial). Also, \begin{align*} \partial_{zz}f&=\frac{1}{2}\left( \partial_{x}\left( \partial_{x}p+\frac{1}{i}\partial_{y}p \right)+\frac{1}{i}\partial_{y}\left( \partial_{x}p+\frac{1}{i}\partial_{y}p \right) \right) \\ &=\frac{1}{2}\left( \partial_{xx}p+2\frac{1}{i}\partial_{xy}p-\partial_{yy}p \right) \end{align*} As before, this says that $\partial_{zz}f$ is a homogeneous polynomial of degree $m-2$. Repeating this argument you have that the $n^{\mathrm{th}}$-derivative of $f$ with respect to $z$ is a homogeneous polynomial of degree $m-n$. Since homogeneous polynomial vanishes at $0$, you obtain the first claim. Note that this does not work for the $m^{\mathrm{th}}$-derivative, because you can have the case that the coefficients of your polynomial (which is possibly the last thing left after taking derivatives) do not sum zero.
I think you are right. Until now you have $f=cz^{m}$. Then $$p=\operatorname{Re}(f)=\frac{cz^{m}+\overline{cz^{m}}}{2}.$$ I think that maybe the authors made some abuse of the notation with the constants in this case... Thus, the authors proved that $$H_{m}(\mathbb{R}^{2}) \subset \{ dz^{m}+\overline{dz^{m}};d \in \mathbb{C}\}.$$ The other inclusion is clear. Therefore, you only have that $$H_{m}(\mathbb{R}^{2})= \{ dz^{m}+\overline{dz^{m}};d \in \mathbb{C}\}, \qquad (*)$$ which is different to have the equality with $\operatorname{span}_{\mathbb{C}}\{z^{m},\overline{z^{m}}\}$. I think that this was just a little error writing what they had in mind, because it makes sense to obtain (*) due to the "symmetry" of the Fourier coefficients: $a_{m}=\overline{a_{-m}}$.