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Complex functions and inner product $\langle \frac{\partial f}{\partial z} , g\rangle $

Complex functions and inner product $\langle \frac{\partial f}{\partial z} , g\rangle $

I'm working through this academic paper : stability of the soljutions of differential equations whose author is Bernard Beauzamy. [A link to paper](https://typeset.io/pdf/stability-of-the-solutions-of-differential-equations-2byjzvdtk5.pdf)
In the academic paper, it work with the next norm
	\begin{equation*}
	\left\Vert f\right\Vert= \left( \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2} |f(re^{i\theta})|^2 r dr\frac{d\theta}{\pi} \right)^{1/2}
	\end{equation*}
for analytic polynomials 
and the scalar product associated is 
	\begin{equation*}
		<f,g>= \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2}f(re^{i\theta})\overline{g(re^{i\theta})} r dr\frac{d\theta}{\pi}
	\end{equation*}
The lemma that i'm trying to check say the following 

For any f,g and $\mathcal{P}_2$ 
	\begin{equation*}
			\langle \frac{\partial f}{\partial z} , g\rangle =\langle f, zg \rangle \qquad \langle zf,g \rangle =\langle f, \frac{\partial g}{\partial z} \rangle
	\end{equation*}
the functions  f  and  g  are given by
	\begin{equation*}
	f(z) = \sum_{j=0}^{\infty} a_j z^j, \quad g(z) = \sum_{j=0}^{\infty} b_j z^j.
	\end{equation*}
The derivative of f with respect to z is
	\begin{equation*}
	\frac{\partial f}{\partial z} = \sum_{j=1}^{\infty} a_j j z^{j-1}.
	\end{equation*}
so the scalar product is
	\begin{align*}
	\langle \frac{\partial f}{\partial z}, g \rangle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=1}^{\infty} a_j j (re^{i\theta})^{j-1}\right) \overline{\left(\sum_{k=0}^{\infty} b_k (re^{i\theta})^k\right)} r \, dr \, \frac{d\theta}{\pi} \\
	&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j j r^{j-1} e^{i(j-1)\theta} \overline{b_k} r^k e^{-ik\theta} r \, dr \, \frac{d\theta}{\pi} \\
	&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} j r^{j+k} e^{i(j-k-1)\theta} r \, dr \, \frac{d\theta}{\pi}.
	\end{align*}

\begin{equation*}
\int_{0}^{2\pi} e^{i(j-k-1)\theta} \frac{d\theta}{\pi} = \begin{cases}
1 & \text{if } j = k+1, \\
0 & \text{if } j \neq k+1.
\end{cases}
\end{equation*}
Hence
\begin{align*}
\langle \frac{\partial f}{\partial z}, g \rangle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) r^{2k+3} r \, dr \\
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}

 The other scalar product is
\begin{align*}
\langle f, zg \rangle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=0}^{\infty} a_j (re^{i\theta})^j\right) \overline{\left(\sum_{k=0}^{\infty} b_k r^{k+1} e^{i(k+1)\theta}\right)} r \, dr \, \frac{d\theta}{\pi} \\
&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} r^{j+k+1} e^{i(j-k-1)\theta} r \, dr \, \frac{d\theta}{\pi}.
\end{align*}


Again, integrating over $ \theta $ and j=k+1 
\begin{align*}
\langle f, zg \rangle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} r^{2k+3} r \, dr \\
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}


\begin{align*}
\langle \frac{\partial f}{\partial z}, g \rangle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr, \\
\langle f, zg \rangle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
These are not equal due to the factor (k+1) present in $\langle \frac{\partial f}{\partial z}, g \rangle $. So i don't know where is the fail.