I'm working through this academic paper: Stability of the Solutions of Differential Equations whose author is Bernard Beauzamy. A link to paper. In the academic paper, it works with the 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*} \langle f,g\rangle= \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 says the following:

For any $f$, $g$ in $\mathcal{P}_2$ \begin{equation*} \left\langle \frac{\partial f}{\partial z} , g\right\rangle = \langle f, zg \rangle \qquad \langle zf,g \rangle = \left\langle f, \frac{\partial g}{\partial z} \right\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*} \left\langle \frac{\partial f}{\partial z}, g \right\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*} \left\langle \frac{\partial f}{\partial z}, g \right\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 $\left\langle \frac{\partial f}{\partial z}, g \right\rangle$. So I don't know where the error is.