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Suppose $f$ is an analytic radial function. I just read in a paper that the number of zeros of $f$ in the interval $[-R,R]$ will then be bounded above by a multiple of $R$. I'm struggling to see how I would go about proving this. Any tips?

Ted
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    Is there any non-constant radial analytic function? – geetha290krm Oct 05 '22 at 11:49
  • The function in question (in this case) is the Fourier transform of the indicator function for a ball, say the unit ball, in $\mathbb{R}^n$. This will be an analytic function, and a radial function. And supposedly, these two things alone should enable me to conclude that the number of zeros for this function in $[-R,R]$ is bounded by $CR$, where presumably $C$ does not depend on $R$. – Ted Oct 07 '22 at 08:16
  • I don't know what notion of analyticity you are using but there is no analytic function in the sense of complex analysis which is radial, except for constants. – geetha290krm Oct 07 '22 at 08:30
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    I don't believe this is meant in the sense of complex analysis, from what I can gather the function should be real analytic. – Ted Oct 07 '22 at 12:52

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