# 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

This is not true. Cauchy's Theorem provides abundant counterexamples.

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 $C(K)$.)

• Can you elaborate on how to construct such measures from Cauchy's theorem, please? – C-Star-W-Star Jun 15 '15 at 16:56
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – TravisJ Jun 15 '15 at 17:26
• @TravisJ Why doesn't this give an answer? Perhaps it should have been written "From Cauchy's theorem it's easy to see there are lots of counterexamples" or something, but it seems like a fine answer to me. – zhw. Jun 15 '15 at 17:59
• @zhw., this could be an answer if the answerer took one or two sentences to explain how cauchy's theorem gives a counter-example. This is more of an assertion that a counter-example exists. – TravisJ Jun 15 '15 at 18:01
• 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