Proving Liouville for entire functions using MVT for analytic functions












1












$begingroup$


I am trying to prove Liouville's theorem: An entire bounded function is constant. I'm trying to use the Mean Value Theorem from my textbook.




MVT: $spacespace $If $f$ is analytic in $D$ and $a in D$, then $f(a)$ is equal to the mean value of $f$ taken around the boundary of any disk centered at $a$ and contained in $D$. That is,


$f(a)$ = $frac{1}{2pi} $$int_{0}^{2pi} f(a + re^{itheta}) dtheta$ when $D(a; r) subset D$



Proof:




Assumtion: Suppose that $f(z)$ is analytic and NOT constant on a circle $C$.


Clearly $f(z)$ is bounded, giving us the necessary assumption (hold your horses for the entire part).




$$ (*) space space space space space spacespace space space|f(z)| leq{frac{1}{2pi}int_{0}^{2pi}|f(z + re^{itheta})| dtheta} leq{max{_theta(|f(z + re^{itheta})}}|)$$

$(**)$ Since $f(z)$ is bounded, $exists{}$ $z$, s.t.equality is achieved in $(*)$



The only way equality holds, is if $f$ is constant throughout the circle. So f is constant throughout the circle



Let $f$ be entire. Since entire functions are by definition holomorphic and analytic on their domain, our statement $(**)$ still holds since $f$ is still bounded by assumption.



So $f$ is constant. Contradiction. $QED$






complex-analysis entire-functions







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$endgroup$












  • $begingroup$
    Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:47


















1












$begingroup$


I am trying to prove Liouville's theorem: An entire bounded function is constant. I'm trying to use the Mean Value Theorem from my textbook.




MVT: $spacespace $If $f$ is analytic in $D$ and $a in D$, then $f(a)$ is equal to the mean value of $f$ taken around the boundary of any disk centered at $a$ and contained in $D$. That is,


$f(a)$ = $frac{1}{2pi} $$int_{0}^{2pi} f(a + re^{itheta}) dtheta$ when $D(a; r) subset D$



Proof:




Assumtion: Suppose that $f(z)$ is analytic and NOT constant on a circle $C$.


Clearly $f(z)$ is bounded, giving us the necessary assumption (hold your horses for the entire part).




$$ (*) space space space space space spacespace space space|f(z)| leq{frac{1}{2pi}int_{0}^{2pi}|f(z + re^{itheta})| dtheta} leq{max{_theta(|f(z + re^{itheta})}}|)$$

$(**)$ Since $f(z)$ is bounded, $exists{}$ $z$, s.t.equality is achieved in $(*)$



The only way equality holds, is if $f$ is constant throughout the circle. So f is constant throughout the circle



Let $f$ be entire. Since entire functions are by definition holomorphic and analytic on their domain, our statement $(**)$ still holds since $f$ is still bounded by assumption.



So $f$ is constant. Contradiction. $QED$






complex-analysis entire-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:47
















1












1








1





$begingroup$


I am trying to prove Liouville's theorem: An entire bounded function is constant. I'm trying to use the Mean Value Theorem from my textbook.




MVT: $spacespace $If $f$ is analytic in $D$ and $a in D$, then $f(a)$ is equal to the mean value of $f$ taken around the boundary of any disk centered at $a$ and contained in $D$. That is,


$f(a)$ = $frac{1}{2pi} $$int_{0}^{2pi} f(a + re^{itheta}) dtheta$ when $D(a; r) subset D$



Proof:




Assumtion: Suppose that $f(z)$ is analytic and NOT constant on a circle $C$.


Clearly $f(z)$ is bounded, giving us the necessary assumption (hold your horses for the entire part).




$$ (*) space space space space space spacespace space space|f(z)| leq{frac{1}{2pi}int_{0}^{2pi}|f(z + re^{itheta})| dtheta} leq{max{_theta(|f(z + re^{itheta})}}|)$$

$(**)$ Since $f(z)$ is bounded, $exists{}$ $z$, s.t.equality is achieved in $(*)$



The only way equality holds, is if $f$ is constant throughout the circle. So f is constant throughout the circle



Let $f$ be entire. Since entire functions are by definition holomorphic and analytic on their domain, our statement $(**)$ still holds since $f$ is still bounded by assumption.



So $f$ is constant. Contradiction. $QED$






complex-analysis entire-functions







share|cite|improve this question











$endgroup$




I am trying to prove Liouville's theorem: An entire bounded function is constant. I'm trying to use the Mean Value Theorem from my textbook.




MVT: $spacespace $If $f$ is analytic in $D$ and $a in D$, then $f(a)$ is equal to the mean value of $f$ taken around the boundary of any disk centered at $a$ and contained in $D$. That is,


$f(a)$ = $frac{1}{2pi} $$int_{0}^{2pi} f(a + re^{itheta}) dtheta$ when $D(a; r) subset D$



Proof:




Assumtion: Suppose that $f(z)$ is analytic and NOT constant on a circle $C$.


Clearly $f(z)$ is bounded, giving us the necessary assumption (hold your horses for the entire part).




$$ (*) space space space space space spacespace space space|f(z)| leq{frac{1}{2pi}int_{0}^{2pi}|f(z + re^{itheta})| dtheta} leq{max{_theta(|f(z + re^{itheta})}}|)$$

$(**)$ Since $f(z)$ is bounded, $exists{}$ $z$, s.t.equality is achieved in $(*)$



The only way equality holds, is if $f$ is constant throughout the circle. So f is constant throughout the circle



Let $f$ be entire. Since entire functions are by definition holomorphic and analytic on their domain, our statement $(**)$ still holds since $f$ is still bounded by assumption.



So $f$ is constant. Contradiction. $QED$






complex-analysis entire-functions




complex-analysis entire-functions






share|cite|improve this question















share|cite|improve this question













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edited Dec 17 '18 at 4:49







lf2225

















asked Dec 17 '18 at 4:18









lf2225lf2225

62




62












  • $begingroup$
    Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:47




















  • $begingroup$
    Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:47


















$begingroup$
Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
$endgroup$
– lf2225
Dec 17 '18 at 4:47






$begingroup$
Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $infty$
$endgroup$
– lf2225
Dec 17 '18 at 4:47












1 Answer
1






active

oldest

votes


















0












$begingroup$

It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:40












  • $begingroup$
    No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
    $endgroup$
    – LutzL
    Dec 17 '18 at 17:40










  • $begingroup$
    What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
    $endgroup$
    – lf2225
    Dec 17 '18 at 19:39













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:40












  • $begingroup$
    No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
    $endgroup$
    – LutzL
    Dec 17 '18 at 17:40










  • $begingroup$
    What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
    $endgroup$
    – lf2225
    Dec 17 '18 at 19:39


















0












$begingroup$

It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:40












  • $begingroup$
    No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
    $endgroup$
    – LutzL
    Dec 17 '18 at 17:40










  • $begingroup$
    What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
    $endgroup$
    – lf2225
    Dec 17 '18 at 19:39
















0












0








0





$begingroup$

It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).






share|cite|improve this answer









$endgroup$



It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 4:28









Robert IsraelRobert Israel

324k23214468




324k23214468












  • $begingroup$
    Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:40












  • $begingroup$
    No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
    $endgroup$
    – LutzL
    Dec 17 '18 at 17:40










  • $begingroup$
    What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
    $endgroup$
    – lf2225
    Dec 17 '18 at 19:39




















  • $begingroup$
    Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
    $endgroup$
    – lf2225
    Dec 17 '18 at 4:40












  • $begingroup$
    No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
    $endgroup$
    – LutzL
    Dec 17 '18 at 17:40










  • $begingroup$
    What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
    $endgroup$
    – lf2225
    Dec 17 '18 at 19:39


















$begingroup$
Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
$endgroup$
– lf2225
Dec 17 '18 at 4:40






$begingroup$
Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree?
$endgroup$
– lf2225
Dec 17 '18 at 4:40














$begingroup$
No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
$endgroup$
– LutzL
Dec 17 '18 at 17:40




$begingroup$
No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle.
$endgroup$
– LutzL
Dec 17 '18 at 17:40












$begingroup$
What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
$endgroup$
– lf2225
Dec 17 '18 at 19:39






$begingroup$
What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $infty$)
$endgroup$
– lf2225
Dec 17 '18 at 19:39




















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