Analytic function on a compact set

Asked Dec 18 '13 at 18:54

Active Dec 18 '13 at 19:15

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I am trying to prove the following but I am not sure how to use the compactness:

Let $f$ be a real analytic function on a compact set $K$. Let $(x_n)_n$ be a sequence in $K$ such that $f(x_n) = 0$. Prove that $f = 0$.

I would appreciate a hint. Thank you!

edited Dec 18 '13 at 19:15

Asaf Karagila

asked Dec 18 '13 at 18:54

Whats My Name

What dimension has the ambient space? $K \subset \mathbb{R}^d$ for which $d$? – Daniel Fischer Dec 18 '13 at 18:56
$R$ is in 1D. real numbers. – Whats My Name Dec 18 '13 at 18:57
Hint, Since $K$ is compact, it is also sequentially compact. – hmakholm left over Monica Dec 18 '13 at 18:57
But, you must require that the sequence $(x_n)$ has infinitely many distinct terms. – Daniel Fischer Dec 18 '13 at 18:57
I hope $K$ is connected; otherwise the statement is false; consider $$ f(x)=\begin{cases}0 & x\in[0,1] \ x & x\in[2,3]\end{cases}$$ and $x_n=1/n$. – hmakholm left over Monica Dec 18 '13 at 18:58
yes I know that and am trying to use the expression of f as an infinite sum around the limit of a converging subsequence of x_n but I am not sure how to prove that that series goes to 0 – Whats My Name Dec 18 '13 at 18:58
if K is in R 1D and it is compact I assume it is an interval right? – Whats My Name Dec 18 '13 at 18:59
@WhatsMyName: How is $[0,1]\cup[2,3]$ not compact? – hmakholm left over Monica Dec 18 '13 at 19:00
Indeed I was in a hurry but yes K is connected. – Whats My Name Dec 18 '13 at 19:02
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This has been asked here before. See this answer, for example. – Andrés E. Caicedo Dec 18 '13 at 19:08
@AndresCaicedo: Thank you! – Whats My Name Dec 18 '13 at 19:24

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