-1
$\begingroup$

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?

  • 0
    uh... I get "page 156 to 176 are not shown in this preview"2011-07-15
  • 1
    Google Book links are unstable and country-dependent. I cannot even see page 174 from where I am connecting right now.2011-07-15
  • 3
    Victor: I urge you to improve the quality of your questions. Try to make them *[self-contained](http://math.stackexchange.com/questions/48742/please-prove-laplaces-theorem-by-english-word-for-word-i-dont-know-where-to-sta)* and *[understandable](http://math.stackexchange.com/questions/51464/please-give-me-a-counter-example-and-example-of-bounded-entire-function-that-is)*, please.2011-07-15
  • 0
    You have also posted it [here](http://www.artofproblemsolving.com/Forum/viewtopic.php?t=418421&p2360294#p2360294).2011-07-15

2 Answers 2

2

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 $\epsilon=1/2$.

1

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| then $|H(z)-H(z_0)|<\varepsilon$. Plug in $z_0=0$, $\varepsilon=\frac{1}{2}$ and remember that $H(0)=0$.