# Complex Measures: Polynomials

Given the complex plane $\mathbb{C}$.

Consider a complex measure: $$\mu:\mathcal{B}(\mathbb{C})\to\mathbb{C}:\quad\operatorname{supp}\mu\subseteq\overline{B_r}$$

Then one has: $$\int\lambda^k\,\mathrm{d}\mu(\lambda)=0\quad(k\in\mathbb{N}_0)\implies\mu=0$$

How can I prove this?

• This question shows us the difference between \mathrm{supp}\mu and \operatorname{supp}\mu : $\mathrm{supp}\mu$ verusus $\operatorname{supp}\mu$. The spacing to the left or right is added only when something is there --- in this case the letter $\mu$ to the right. I edited accordingly. – Michael Hardy Jun 15 '15 at 17:40
• Thanks @MichaelHardy!! (I prefer the one without spacing...) – C-Star-W-Star Jun 15 '15 at 18:10
• Note that the implication is true for: $\int\lambda^k\overline{\lambda}^l\,\mathrm{d}\mu(\lambda)=0\quad(k,l\in\mathbb{N}_0)\implies\mu=0$ – C-Star-W-Star Jun 16 '15 at 23:11
• That implication is true if $|mu$ has compact support. In general there's no reason the integrals in your original post or the integrals in your comment should even exist. – David C. Ullrich Jun 22 '15 at 15:19
• @DavidC.Ullrich: Yep, right! Made a quick remark to keep record but forgot to mention compact support. I guess I was just distracted. I was quite impressed by the fact that such measures exist at all. But then I realized it is due to Cauchy's theorem as you mentioned first. Btw, do you mind adding that key remark in short to your answer. Hope I didn't put to much anger lately. – C-Star-W-Star Jun 22 '15 at 15:30

In detail: Suppose $\gamma$ is a smooth closed curve in the plane. Cauchy's Theorem says that if $f$ is entire then $\int_\gamma f(z)\,dz=0$. But it's clear that there exists a complex measure $\mu$ such that $$\int f\,d\mu=\int_\gamma f(z)\,dz.\quad(*)$$
(Readers to whom the existence of $\mu$ is not clear are advised to contemplate the Riesz Representation Theorem, describing the dual of $C(K)$.)
• Do we know Cauchy's Theorem? Special case with simple hypotheses: If $f$ is entire and $\gamma$ is a smooth closed curve then $$\int_\gamma f(z)\,dz=0.$$ Surely it's clear that given $\gamma$ there exists $\mu$ so that $\int_\gamma f dz=\int f\,d\mu$? – David C. Ullrich Jun 15 '15 at 20:43