How to prove that $f(z)$ is bounded if we know that $limlimits_{|z|to infty}frac{f(z)}{z}= 0$?
-1
$begingroup$
Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.
How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$
complex-analysis functional-analysis complex-numbers
edited Dec 27 '18 at 21:44
user376343
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
$endgroup$
add a comment |
-1
$begingroup$
Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.
How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$
complex-analysis functional-analysis complex-numbers
edited Dec 27 '18 at 21:44
user376343
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
$endgroup$
2
$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04
add a comment |
-1
-1
-1
$begingroup$
Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.
How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$
complex-analysis functional-analysis complex-numbers
edited Dec 27 '18 at 21:44
user376343
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
$endgroup$
Knowing that $f(z)$ is analytic on the entire complex plane and that $limlimits_{|z| to infty} frac{f(z)}{z} = 0$.
How do I prove that $f(z)$ have is bounded ? Meaning there exists a real number $M$ such that for every $z, |f(z)|< M.$
complex-analysis functional-analysis complex-numbers
complex-analysis functional-analysis complex-numbers
edited Dec 27 '18 at 21:44
user376343
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
edited Dec 27 '18 at 21:44
user376343
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
edited Dec 27 '18 at 21:44
user376343
3,8483829
edited Dec 27 '18 at 21:44
user376343
3,8483829
edited Dec 27 '18 at 21:44
user376343
3,8483829
3,8483829
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
asked Dec 23 '18 at 19:22
Idan AvivIdan Aviv
1014
1014
2
$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04
add a comment |
2
$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04
2
2
$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04
add a comment |
2 Answers
2
active
oldest
votes
3
$begingroup$
If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
1
$begingroup$
Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$
Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.
Finally,
$$0= lim_{z to infty} g(z)= C$$
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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3
$begingroup$
If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
3
$begingroup$
If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
3
3
3
$begingroup$
If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
If $ninmathbb N$, then$$f^{(n)}(0)=frac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dz$$and therefore $f^{(n)}(0)=0$ if $n>1$. So, there are $a,binmathbb C$ such that $(forall zinmathbb{C}):f(z)=az+b$. But, since $lim_{ztoinfty}frac{f(z)}z=0$, $a=0$. So, $f$ is actually constant.
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
edited Dec 23 '18 at 19:53
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 23 '18 at 19:32
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
$endgroup$
– Idan Aviv
Dec 23 '18 at 19:38
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
Typo. I meant $n>1$. I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 19:54
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
I have two questions. 1. How do you know that for n>1 the n derivative of f(z) is 0? And why would n = 1 be different from zero ? 2. How do you know that f(z) is a polynomial?
$endgroup$
– Idan Aviv
Dec 23 '18 at 20:17
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
1. $leftlvertfrac1{2pi i}oint_{partial D(0,R)}frac{f(z)}{z^{n+1}},mathrm dzrightrvertleqslantsup_{lvert zrvert=R}frac{Rlvert f(z)rvert}{R^{n+1}}sup_{lvert zrvert=R}frac{lvert f(z)rvert}{R^n}to0$. 2. Since $f^{(n)}(0)=0$ when $n>1$ and $f$ is entire, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 21:18
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
$begingroup$
One last question, wouldn't the limit as R approach infinity will be zero when n = 1. Because in your answer you wrote that this limit will approach zero for n > 1, and then I need to do further check in order to show that the first derivative is zero. So why is the limit not true for n=1?
$endgroup$
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
1
$begingroup$
Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$
Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.
Finally,
$$0= lim_{z to infty} g(z)= C$$
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
$endgroup$
add a comment |
1
$begingroup$
Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$
Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.
Finally,
$$0= lim_{z to infty} g(z)= C$$
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
$endgroup$
add a comment |
1
1
1
$begingroup$
Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$
Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.
Finally,
$$0= lim_{z to infty} g(z)= C$$
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
$endgroup$
Hint If $f(z)$ is entire, so is
$$g(z)=frac{f(z)-f(0)}{z}$$
Next, it is easy to see that
$$lim_{z to infty}g(z)=0$$
which implies that $g$ is entire and bounded, thus constant. There fore, the exists some $C$ such that $g(z)=C forall x$.
Finally,
$$0= lim_{z to infty} g(z)= C$$
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
answered Dec 27 '18 at 21:55
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
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$begingroup$
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 19:27
$begingroup$
Another way is to look at the Laurent Series of $t f(1/t)$ at $t=0$. The above relationship clearly isn't held for non-constant polynomials, so if $f$ isn't constant, then it must be an entire non-polynomial, so it must have an essential singularity at $ z = infty$. But we know by the Laurent series $f(z)/z$ is also non-polynomial, so $lim_{z rightarrow infty} f/z = 0$ contradicts that the image of $f(z)/z$ is dense in $mathbb{C}$ near $infty$, so $f$ must be constant.
$endgroup$
– Story123
Dec 23 '18 at 21:04