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$f$ is analytic function on open unit disk, and $|f(z)|\le 1-|z|\forall z\in D$, we need to show $f\equiv 0$, just a hint please.

Myshkin
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1 Answers1

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What value does $1-|z|$ take on $\partial D$? Then apply the maximum modulus theorem appropriately.

Answer:

Let $ r \in (0,1)$, and consider $f$ on $B(0,r)$. Note that $\partial B(0,r) = \{z \, | \, |z|=r\}$, and hence $|f(z)| \leq 1-|z| = 1-r$ for $z \in \partial B(0,r)$. By the maximum modulus theorem, this gives $|f(z)| \leq 1-r$ for $z \in B(0,r)$.

Now fix $z \in D$ (hence $|z|<1$). Note that $z \in B(0,r)$ for all $r \in (|z|,1)$. Hence it follows from above that $|f(z)| \leq 1-r$ for all $r \in (|z|,1)$. Hence $f(z) = 0$.

copper.hat
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  • $g(z)=1-|z|$ takes $0$ on $\partial D$ but my $f$ is defined on open unit disk. – Myshkin Jan 27 '13 at 06:42
  • Did you want a hint or a solution? Try looking at the value of $f$ on $B(r,0)$, with $r \in (0,1)$. – copper.hat Jan 27 '13 at 06:44
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    Although $f$ is defined on the open disk, it has a continuous extension to the closed disk. – Robert Israel Jan 27 '13 at 06:49
  • Alternatively, but this requires a little more machinery, you could note that $f$ can be extended to a continuous function on $\overline{D}$. Then $f$ is analytic on $D$, continuous on $\overline{D}$ (in fact, $f(z) = 0$ on $\partial D$). Then apply the Poisson integral formula. – copper.hat Jan 27 '13 at 06:52
  • I dont know about poisson integral,well, $f$ is continuous on $\bar{D}$ hence attain maximum some where in $\bar{D}$, but as $f$ is analytic so it must attain maxima on the boundary, but on the boundary it takes values $0$, so max$|f|=0$?, please tell me detail solution if i am getting wrong. – Myshkin Jan 27 '13 at 06:59
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    I have added the solution above. – copper.hat Jan 27 '13 at 07:07
  • I have not understood at all.please write some words – Myshkin Jan 27 '13 at 07:45
  • I have added more words. If it is not clear, please be explicit about what it is that needs to be clarified. – copper.hat Jan 27 '13 at 08:06
  • everything is fine but just the last line, why are you claimining $f(z)=0$, because $|z|=1$ never, is there any $r\rightarrow 1$ business inside?, if yes then okay.thank you very much for your help. – Myshkin Jan 27 '13 at 12:43
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    Well, I have shown that for an $z \in D$, you have $|f(z)| \leq 1-r$ for any $r \in (|z|,1)$. The only way this can be true is if $f(z) = 0$ (otherwise, choose $r > 1-\frac{1}{2}|f(z)|$, which gives $0<|f(z)| \leq \frac{1}{2}|f(z)|$, a contradiction). – copper.hat Jan 27 '13 at 14:59