Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

please see here(p.174-175 Elementary real and complex analysis By Georgiĭ Evgenʹevich Shilov):

Image snapshot from google books:

enter image description here

Question is, why is $\displaystyle |H(z)| \lt 1/2$ true?

share|cite|improve this question
uh... I get "page 156 to 176 are not shown in this preview" – leonbloy Jul 15 '11 at 18:30
Google Book links are unstable and country-dependent. I cannot even see page 174 from where I am connecting right now. – Arturo Magidin Jul 15 '11 at 18:32
Victor: I urge you to improve the quality of your questions. Try to make them self-contained and understandable, please. – t.b. Jul 15 '11 at 18:35
You have also posted it here. – Ehsan M. Kermani Jul 15 '11 at 18:50
up vote 2 down vote accepted

It is the definition of continuity in $0$. For any $\epsilon$ you can find $\delta$ such as $z$ lies in the disc centered in $0$ with radius $\delta$ implies $H(z)$ is in the disc centered on $H(0)=0$ with radius $\epsilon=1/2$.

share|cite|improve this answer

It follows from continuity of polynomials: If $H$ is continuus at $z_0$ then for any $\varepsilon>0$ there exists $r>0$ such that if $|z-z_0|<r$ then $|H(z)-H(z_0)|<\varepsilon$. Plug in $z_0=0$, $\varepsilon=\frac{1}{2}$ and remember that $H(0)=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.