An Application of the Phragmen-Lindelof Theorem












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This is Chapter 6, section 4, exercise 7 (page 141) from Functions of One Complex Variable by John B. Conway, Second Edition.



Exercise 6.4.7: Let $G={z:Re (z)>0}$ and let $f:Grightarrowmathbb{C}$ be analytic such that $f(1)=0$ and such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for $w$ in $partial G$. Also suppose that for every $delta, 0<delta<1,$ there is a constant $P$ such that $$|f(z)|leq Pexpleft(|z|^{1-delta}right)$$ Prove that $$|f(z)|leq Mleft[dfrac{(1-x)^2+y^2}{(1+x)^2+y^2}right]^{1/2}$$ Hint: Consider $f(z)left(dfrac{1+z}{1-z}right)$



I want to use Corollary 4.2 in this section which says: Let $ageqdfrac{1}{2}$ and put $$G=left{z:|arg z|<dfrac{pi}{2a}right}$$ Suppose that $f$ is analytic on $G$ and there is a constant $M$ such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for all $w$ in $partial G$. If there are positive constants $P$ and $b<a$ such that $$|f(z)|leq Pexpleft(|z|^bright)$$ for all $z$ with $|z|$ sufficiently large, then $|f(z)|leq M$ for all $z$ in $G$.










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$endgroup$












  • $begingroup$
    Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 5:48












  • $begingroup$
    How do I apply the corollary?
    $endgroup$
    – 111
    Dec 15 '18 at 8:19
















0












$begingroup$


This is Chapter 6, section 4, exercise 7 (page 141) from Functions of One Complex Variable by John B. Conway, Second Edition.



Exercise 6.4.7: Let $G={z:Re (z)>0}$ and let $f:Grightarrowmathbb{C}$ be analytic such that $f(1)=0$ and such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for $w$ in $partial G$. Also suppose that for every $delta, 0<delta<1,$ there is a constant $P$ such that $$|f(z)|leq Pexpleft(|z|^{1-delta}right)$$ Prove that $$|f(z)|leq Mleft[dfrac{(1-x)^2+y^2}{(1+x)^2+y^2}right]^{1/2}$$ Hint: Consider $f(z)left(dfrac{1+z}{1-z}right)$



I want to use Corollary 4.2 in this section which says: Let $ageqdfrac{1}{2}$ and put $$G=left{z:|arg z|<dfrac{pi}{2a}right}$$ Suppose that $f$ is analytic on $G$ and there is a constant $M$ such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for all $w$ in $partial G$. If there are positive constants $P$ and $b<a$ such that $$|f(z)|leq Pexpleft(|z|^bright)$$ for all $z$ with $|z|$ sufficiently large, then $|f(z)|leq M$ for all $z$ in $G$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 5:48












  • $begingroup$
    How do I apply the corollary?
    $endgroup$
    – 111
    Dec 15 '18 at 8:19














0












0








0


1



$begingroup$


This is Chapter 6, section 4, exercise 7 (page 141) from Functions of One Complex Variable by John B. Conway, Second Edition.



Exercise 6.4.7: Let $G={z:Re (z)>0}$ and let $f:Grightarrowmathbb{C}$ be analytic such that $f(1)=0$ and such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for $w$ in $partial G$. Also suppose that for every $delta, 0<delta<1,$ there is a constant $P$ such that $$|f(z)|leq Pexpleft(|z|^{1-delta}right)$$ Prove that $$|f(z)|leq Mleft[dfrac{(1-x)^2+y^2}{(1+x)^2+y^2}right]^{1/2}$$ Hint: Consider $f(z)left(dfrac{1+z}{1-z}right)$



I want to use Corollary 4.2 in this section which says: Let $ageqdfrac{1}{2}$ and put $$G=left{z:|arg z|<dfrac{pi}{2a}right}$$ Suppose that $f$ is analytic on $G$ and there is a constant $M$ such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for all $w$ in $partial G$. If there are positive constants $P$ and $b<a$ such that $$|f(z)|leq Pexpleft(|z|^bright)$$ for all $z$ with $|z|$ sufficiently large, then $|f(z)|leq M$ for all $z$ in $G$.










share|cite|improve this question











$endgroup$




This is Chapter 6, section 4, exercise 7 (page 141) from Functions of One Complex Variable by John B. Conway, Second Edition.



Exercise 6.4.7: Let $G={z:Re (z)>0}$ and let $f:Grightarrowmathbb{C}$ be analytic such that $f(1)=0$ and such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for $w$ in $partial G$. Also suppose that for every $delta, 0<delta<1,$ there is a constant $P$ such that $$|f(z)|leq Pexpleft(|z|^{1-delta}right)$$ Prove that $$|f(z)|leq Mleft[dfrac{(1-x)^2+y^2}{(1+x)^2+y^2}right]^{1/2}$$ Hint: Consider $f(z)left(dfrac{1+z}{1-z}right)$



I want to use Corollary 4.2 in this section which says: Let $ageqdfrac{1}{2}$ and put $$G=left{z:|arg z|<dfrac{pi}{2a}right}$$ Suppose that $f$ is analytic on $G$ and there is a constant $M$ such that $limsuplimits_{zrightarrow w}|f(z)|leq M$ for all $w$ in $partial G$. If there are positive constants $P$ and $b<a$ such that $$|f(z)|leq Pexpleft(|z|^bright)$$ for all $z$ with $|z|$ sufficiently large, then $|f(z)|leq M$ for all $z$ in $G$.







complex-analysis analysis complex-numbers upper-lower-bounds






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edited Dec 15 '18 at 23:41







111

















asked Dec 15 '18 at 5:39









111111

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  • $begingroup$
    Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 5:48












  • $begingroup$
    How do I apply the corollary?
    $endgroup$
    – 111
    Dec 15 '18 at 8:19


















  • $begingroup$
    Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 5:48












  • $begingroup$
    How do I apply the corollary?
    $endgroup$
    – 111
    Dec 15 '18 at 8:19
















$begingroup$
Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 5:48






$begingroup$
Does $z=x+iy$? If so, your last equation literally reads $$frac{1+z}{1-z}leleft|frac{1-z}{1+z}right|.$$ But this makes no sense; one cannot have a non-real number on one side of an inequality.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 5:48














$begingroup$
How do I apply the corollary?
$endgroup$
– 111
Dec 15 '18 at 8:19




$begingroup$
How do I apply the corollary?
$endgroup$
– 111
Dec 15 '18 at 8:19










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