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

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.

## 2 answers

There are multiple issues with how you compute the exponent of $r$.

The first issue is you compute $r^{j-1}r^k = r^{j+k}$ rather than $r^{j+k-1}$.

You compound this error when you substitute $j = k+1$ into $r^{j+k}$ (which should be $r^{j+k-1}$) in the formula for $\left\langle\frac{\partial f}{\partial z},g\right\rangle$. You end up with $r^{2k+3}$ rather than $r^{2k+1}$ (which should actually be $r^{2k}$).

You make the same mistake in the formula for $\langle f,zg\rangle$ where $r^{j+k+1}$ should become $r^{2k+2}$ but you have $r^{2k+3}$. That said, in this case you make another mistake which cancels this out by "multiplying" $r^{2k+3}r$ to just $r^{2k+3}$ instead of $r^{2k+4}$ when you pull the sum out of the integral.

Without these mistakes you would end up with $$\begin{align} \left\langle \frac{\partial f}{\partial z}, g\right\rangle &= \sum_{k=0}^\infty a_{k+1}\overline{b_k} (k+1)\int_0^\infty e^{-r^2}r^{2k+1} dr \\ \langle f, zg \rangle &= \sum_{k=0}^\infty a_{k+1}\overline{b_k} \int_0^\infty e^{-r^2}r^{2k+3} dr \end{align}$$

Applying integration by parts to the integral in the second equation will reduce it to the first, i.e. a $k+1$ factor will pop out and the power of the $r^{2k+3}$ will be reduced by $2$. (This is easiest to see, for me at least, by first doing a $u$-substitution with $u=-r^2$.)

#### 1 comment thread

The first case $\langle \frac{\partial f}{\partial z} , g \rangle$. Change of variable \begin{align*} u =r^2 \ \qquad du =2r dr \end{align*} \begin{align*} (k+1)\int_{0}^{\infty} e^{-r^2}r^{2k+1}dr=\int_{0}^{\infty} e^{-u}u^{k +1/2}\frac{1}{2\sqrt{u}}du = \frac{1}{2}\int_{0}^{\infty}e^{-u}u^{k}du = (k+1)\frac{\Gamma\left((k+1)/1\right)}{2}=(k+1)\frac{\Gamma (k+1)}{2}=\frac{(k+1)!}{2} \end{align*} The second case $\langle f, zg \rangle $. The same change of variables \begin{align*} (k+1)\int_{0}^{\infty} e^{-r^2}r^{2k+3}dr=\int_{0}^{\infty} e^{-u}u^{k +3/2}\frac{1}{2\sqrt{u}}du = \frac{1}{2}\int_{0}^{\infty}e^{-u}u^{k+1}du = \frac{\Gamma\left((k+2)/1\right)}{2}=\frac{\Gamma (k+2)}{2}=\frac{(k+1)!}{2} \end{align*} Hence, the inner products are equals. (In the last equality i used the gauss integral and its relation with Gamma function. Look at in Wikipedia)

## 2 comment threads