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Let $$\chi_n(z) = \frac{1}{\sqrt{2\pi}}z^{n-1}.$$

Prove $(\chi_n)$ is orthonormal on $\partial B(0,1)$ in regard to $$\langle f(z), g(z)\rangle = \int_{\partial B(0,1)}f(z)\overline{g(z)}\operatorname dS.$$

Choose $\chi_k(z), \chi_j(z)$ then consider:

$$\begin{aligned} \langle \chi_k(z),\chi_j(z)\rangle & = \frac{1}{2\pi}\int_{\partial B(0,1)} z^{k-1} \overline{z^{j-1}}\operatorname d S\\ &= \frac{1}{2\pi}\int_{\partial B(0,1)}z^{k-1}{\overline z}^{j-1} \operatorname d S \end{aligned}$$ New parameterize $z(\theta) = e^{i\theta}$, $\operatorname d z = ie^{i\theta} \operatorname d \theta$ where $\theta: 0 \to 2\pi$, then

$$\begin{aligned} \langle \chi_k(z),\chi_j(z)\rangle & = \frac{i}{2\pi}\int_0^{2\pi}e^{ik\theta-i\theta}e^{-ij\theta+i\theta}e^{i\theta}\operatorname d \theta\\ & = \frac{i}{2\pi}\int_0^{2\pi}e^{i\theta(k-j+1)}\operatorname d \theta \end{aligned}$$

But this doesn't seem right, when $j=k$ this evaluates to zero, where I would want $\langle \chi_k, \chi_j\rangle = \delta_{kj}$

Where am I doing something wrong?

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  • $\begingroup$ I don't think $dS$ and $dz$ are actually the same here: note that the integral of $1$ against $dS$ should be the circumference, while the integral of $1$ against $dz$ is of course zero. $\endgroup$ – Ian Jan 10 '16 at 20:25
  • $\begingroup$ hmm, even with $\operatorname d S = r\operatorname d\theta$ this doesn't seem right... $\endgroup$ – dietervdf Jan 10 '16 at 20:36
  • $\begingroup$ $dS=r d \theta$ is correct, and does the right thing (when $k=j$ you are just integrating $1$). $\endgroup$ – Ian Jan 10 '16 at 20:37
  • $\begingroup$ I don't see it yet, changing $\operatorname dS = r\operatorname d\theta$ would turn the last integral into $$\frac{i}{2\pi}\int_0^{2\pi} e^{i\theta(k-j+1)} r\operatorname d\theta$$, I don't see what changes... $\endgroup$ – dietervdf Jan 10 '16 at 20:41
  • $\begingroup$ No it doesn't: the extra factor of $e^{i \theta}$ right before your first $d \theta$ is only there because of $dz$, it is not there with $dS$. The same is true of the factor of $i$. So you should have $\frac{1}{2 \pi} \int_0^{2 \pi} e^{i \theta(k-j)} r d \theta$ ($r$ is of course $1$ here). $\endgroup$ – Ian Jan 10 '16 at 20:42
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The surface measure $dS$ is not the same as $dz$. In particular, $dS=d \theta$ (here $r=1$), while $dz=i e^{i \theta} d \theta$. Making that correction, your integral is correctly written as

$$\frac{1}{2 \pi} \int_0^{2 \pi} e^{i \theta(k-j)} d \theta.$$

This is readily seen to be $1$ when $k=j$.

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