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Question: Let $f$ be analytic in a neighborhood of $\overline{\mathbb{D}}$ (the closed unit disk) such that $\int_{|z|=1}|f(z)-z||dz|\leq\frac{1}{100}$. Prove that $f$ has at least one zero in $\mathbb{D}$.

Thoughts: I feel like this should use Maximum Modulus principle, but I can't seem to get that argument to work out. I've seen some similar questions, but nothing that gives us an assumption like the integral here. I've thought maybe something like Rouche or Schwarz may help, but I continue to get nowhere. Any help is greatly appreciated! Thank you.

User7238
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    It does feel Rouché-like, doesn't it? Maybe look at the proof (rather than the statement) of Rouché's theorem, to see if the pointwise inequality in that statement can be replaced by the averaged inequality you have here? – Greg Martin Dec 05 '21 at 19:00
  • @GregMartin That is where I was stuck. I wasn't sure if we could do that, or if I was breaking some "big thing" somewhere. – User7238 Dec 05 '21 at 19:21

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Let $f(z)=\sum a_kz^k$.

Note that $\int_{|z|=1}|f(z)-z||dz|=\int_{|z|=1}|f(z)/z^k-1/z^{k-1}||dz| \ge$

$|\int_{|z|=1}(f(z)/z^k-1/z^{k-1})dz|$,

and this last integral is $2\pi |a_{k-1}|$ if $k \ge 1, k \ne 2$ and $2\pi |a_1-1|$ if $k=2$, so one gets $|a_k| \le 1/(200 \pi)$ if $k \ne 1$ and $|a_1| \ge 1-1/(200\pi)$.

But now one can apply Rouche on say $|z|=1/2$ to $f, a_1z$ as a simple majorization shows that $|f-a_1z| <|a_1|/2$ for $|z|=1/2$ and we are done; one can of course do much better than $1/2$ in finding a $c<1$ st $f$ has a zero with $|z|<c$ if needed

Conrad
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  • Hi Conrad! Could you elaborate a bit on your second and third line? I've just never seen that before, so I am not sure if that is a theorem, or if I am just not seeing some simple algebra. – User7238 Dec 05 '21 at 20:08
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    $|z|=1$ so $|g(z)|=|g(z)/z^k|$ for any integer $k$ and any function $g$ there, while the following inequality is just the triangle inequality for integrals $|\int_{\gamma} g(z)dz| \le \int_{\gamma}|g(z)||dz|$ – Conrad Dec 05 '21 at 20:12
  • Ahhhh! I see! That makes sense. Very nice! +1. I haven't accepted the answer, just because I'm curious if anything else comes up, but this is very nice!! – User7238 Dec 05 '21 at 20:15
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    Happy to be of help – Conrad Dec 05 '21 at 20:16
  • On another note, there was a problem you helped me with quite some time ago (dealing with the cross product) that I commented on a couple days ago... if you had time, would you mind taking a peak? – User7238 Dec 05 '21 at 20:17
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    If you can post a link to it here in the comments, definitely – Conrad Dec 05 '21 at 20:18
  • https://math.stackexchange.com/questions/4219342/mapping-the-circle-z-3-into-z-1-1-the-point-33i-into-1-and-the-poi

    I am wondering if I am off when taking my points of symmetry... in particular, the $\bar{z}$ in the $z^{\star}=\frac{R^2}{\bar{z}-\bar{a}}+a$

     – User7238 Dec 05 '21 at 20:21
  • All looks good when u substitute and answers match so maybe u did wrong substitution – Conrad Dec 05 '21 at 20:45