How to prove that $f(z)$ is bounded if we know that $limlimits_{|z|to infty}frac{f(z)}{z}= 0$?
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
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
add a comment |
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
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
2
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
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.
– Story123
Dec 23 '18 at 21:04
add a comment |
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
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
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
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
edited Dec 27 '18 at 21:44
user376343
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
edited Dec 27 '18 at 21:44
user376343
2,8932823
edited Dec 27 '18 at 21:44
user376343
2,8932823
edited Dec 27 '18 at 21:44
user376343
2,8932823
2,8932823
asked Dec 23 '18 at 19:22
Idan Aviv
1014
asked Dec 23 '18 at 19:22
Idan Aviv
1014
asked Dec 23 '18 at 19:22
Idan Aviv
1014
1014
2
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
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.
– Story123
Dec 23 '18 at 21:04
add a comment |
2
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
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.
– Story123
Dec 23 '18 at 21:04
2
2
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
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.
– Story123
Dec 23 '18 at 21:04
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.
– Story123
Dec 23 '18 at 21:04
add a comment |
2 Answers
2
active
oldest
votes
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 Santos
151k22123224
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
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.
102k5110205
add a comment |
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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 Santos
151k22123224
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
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 Santos
151k22123224
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
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 Santos
151k22123224
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 Santos
151k22123224
edited Dec 23 '18 at 19:53
answered Dec 23 '18 at 19:32
José Carlos Santos
151k22123224
answered Dec 23 '18 at 19:32
José Carlos Santos
151k22123224
answered Dec 23 '18 at 19:32
José Carlos Santos
151k22123224
151k22123224
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Thanks for the quick respond @José Carlos Santos, you wrote: "if n>.". what n should be bigger than?
– Idan Aviv
Dec 23 '18 at 19:38
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
Typo. I meant $n>1$. I've edited my answer. Thank you.
– José Carlos Santos
Dec 23 '18 at 19:54
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?
– Idan Aviv
Dec 23 '18 at 20:17
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?
– Idan Aviv
Dec 23 '18 at 20:17
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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, $(forall zinmathbb{C}):f(z)=f(0)+f'(0)z$.
– José Carlos Santos
Dec 23 '18 at 21:18
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?
– Idan Aviv
Dec 23 '18 at 22:12
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?
– Idan Aviv
Dec 23 '18 at 22:12
|
show 2 more comments
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.
102k5110205
add a comment |
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.
102k5110205
add a comment |
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.
102k5110205
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.
102k5110205
answered Dec 27 '18 at 21:55
N. S.
102k5110205
answered Dec 27 '18 at 21:55
N. S.
102k5110205
answered Dec 27 '18 at 21:55
N. S.
102k5110205
102k5110205
add a comment |
add a comment |
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2
Look at the Cauchy integral formula for the derivative and higher-order derivatives of $f$ at $0$.
– Ted Shifrin
Dec 23 '18 at 19:27
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.
– Story123
Dec 23 '18 at 21:04