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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?

C-star-W-star
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    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
  • @DavidC.Ullrich: Besides can you have a look on this proof, please: Spectral Measures: Uniqueness – C-star-W-star Jun 22 '15 at 15:33
  • I'm not sure, but based on the bits of notation that I don't follow it looks like you're talking about unbounded densely defined operators. I know nothing about the spectral theorem for such things. (Uniqueness for bounded operators is fairly simple.) – David C. Ullrich Jun 22 '15 at 15:58
  • @DavidC.Ullrich: Yeah for bounded operators it is rather simple as everything goes in the realm of bounded measurables. Hmm, ok. Do you maybe know someone here for that on MSE? – C-star-W-star Jun 22 '15 at 18:00

1 Answers1

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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 $C(K)$.)

David C. Ullrich
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    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
  • Maybe not. GIven a smooth curve $\gamma$ define $$\mu(E)=\int_\gamma\chi_E(z),dz.$$ – David C. Ullrich Jun 15 '15 at 21:01
  • Come to think of it, we also need the theorem that if $A=B$ and $B=0$ then $A=0$. I really should have been more clear, sorry. – David C. Ullrich Jun 15 '15 at 21:15
  • Can you outline the construction of the measure little more, please? I get for example: $\int_\gamma f(z)\mathrm{d}z=\int_0^1f(\gamma(s))\dot{\gamma}(s)\mathrm{d}s=\int_0^1f(\gamma (s))\mathrm{d}\mu_\dot{\gamma}(s)$; $\mu_\dot{\gamma}(A):=\int_A\dot{\gamma}(s)\mathrm{d}s=\int_0^1\chi_A(s)\dot{ \gamma}(s)\mathrm{d}s$, $A\subseteq[0,1]$; $\int_\gamma\chi_E(z)\mathrm{d}z=\int_0^1\chi_E(\gamma(s))\dot{\gamma}(s)\mathrm{d}s$, $E\subseteq\mathbb{C}$ – C-star-W-star Jun 16 '15 at 13:19
  • Hmm. My advice, for what it's worth, is to review the material in that course on measure theory - this is standard (basic, important) stuff. Tell me which part you don't get: (i) If we define $\mu(E)=\int_\gamma\chi_E(z)dz$ then $\mu$ is a complex Borel measure on $\mathbb{C}$. (ii) If we define $\mu$ as in (i) then it follows that $\int f,d\mu=int_\gamma f,dz$. (i) Dominated convergence shows that $\mu$ is countably additive, qed. (ii): (i) says that (ii) holds for $f=\chi_E$. Hence (ii) holds for simple functions, and hence for, say, bounded Borel functions $f$ by the usual approximations. – David C. Ullrich Jun 16 '15 at 14:38
  • Got it! ;) Do you mind if I expand your answer? (So it is recorder.) – C-star-W-star Jun 16 '15 at 21:37
  • Oh but problem: Entire functions are not bounded!! – C-star-W-star Jun 16 '15 at 21:41
  • Expanding the answer would be silly. The version of the answer of the proper length really is the original "Cauchy's Theorem gives counterexamples". No, entire functions are not bounded. But they're bounded on the support of $\mu$. – David C. Ullrich Jun 16 '15 at 21:52
  • Sorry for expanding your answer!! :/ I hope this one is fine. Besides I've just read the introductory chapter of your complex analysis book. I have to admit it is just excellent!!! I never have read such a perfect treatment!!! Even the mention of Frechet derivative. Impressed. – C-star-W-star Jun 22 '15 at 11:33
  • Ok. You said in that other thread you wouldn't do it again, and you figured out all by yourself that maybe something should be changed here, fine. So I'll stop shouting. But it seems like you don't get it. You said you wouldn't do it again, but you're still doing it! None of that post was written by me, but it has my name on it. It's like reverse plagiarism or something... (None of it was written by me? I count 35 lines. 34 of them are nothing whatever like anything I wrote. The first sentence is close to something I wrote. But you added the italics and the exclamation point.) – David C. Ullrich Jun 22 '15 at 15:28
  • Maybe this is part of what you don't get: One can easily imagine two versions of the same proof, one a single paragraph and one that takes ten pages. Some readers will not be able to understand the one-paragraph version. But many others will find the one-paragraph version much simpler and clearer! Now, when I write a one-paragraph proof you can ask about details you don't get. You can wish I'd included more details. You can suggest I might want to edit, adding details. But simply going ahead and changing my version to yours, but keeping it under my name, is just not right. – David C. Ullrich Jun 22 '15 at 15:36
  • I simply didn't write that thing that you're insisting on telling the world I wrote. And it's not just a detail or two, your version is like ten times longer than mine! So I'm going to put it back to what I think it should be. You can take your version and post it here or wherever you want. But your version should have your name on it, not mine. – David C. Ullrich Jun 22 '15 at 15:38
  • See, the first sentence "Cauchy's Theorem provides abundant counterexamples" really is sufficient for the vast majority of readers who know enough analysis to be interested in the question in the first place. The part after "In detail" really is just adding details that should be clear to most readers in the intended audience. – David C. Ullrich Jun 22 '15 at 15:49