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 $G$.
complex-analysis analysis complex-numbers upper-lower-bounds
asked Dec 15 '18 at 5:39
111111
275
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add a comment |
$begingroup$
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 $G$.
complex-analysis analysis complex-numbers upper-lower-bounds
asked Dec 15 '18 at 5:39
111111
275
$endgroup$
$begingroup$
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.
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– Lord Shark the Unknown
Dec 15 '18 at 5:48
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How do I apply the corollary?
$endgroup$
– 111
Dec 15 '18 at 8:19
add a comment |
$begingroup$
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 $G$.
complex-analysis analysis complex-numbers upper-lower-bounds
asked Dec 15 '18 at 5:39
111111
275
$endgroup$
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
complex-analysis analysis complex-numbers upper-lower-bounds
asked Dec 15 '18 at 5:39
111111
275
asked Dec 15 '18 at 5:39
111111
275
edited Dec 15 '18 at 23:41
111
asked Dec 15 '18 at 5:39
111111
275
asked Dec 15 '18 at 5:39
111111
275
asked Dec 15 '18 at 5:39
111111
275
275
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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