Mixing
If an entire function grows slower than a polynomial, then it is a polynomial!
$begingroup$
I was investigating the following corollary to Liouville's Theorem in Complex Analysis:
if $f(z)$ is entire and $lim_{zrightarrow infty}z^{-n}f(z)=0$, then $f(z)$ is a polynomial in $z$ of degree at most $n-1$.
My proof uses the complex L'Hospital Rule and Induction on $n$. The base case $n=0$ is obvious. To illustrate the idea: for the case $n=2$ Suppose that $f(z)$ is not bounded, so that $$lim_{zrightarrow infty}{f(z)}=infty.$$ Then by using L'Hospital, we have $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}.$$ If $f'(z)$ is not bounded, then we have further $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}=lim_{zrightarrow infty}{frac{f''(z)}{2}}=0.$$ Thus, by Liouville's Theorem, we know that $f''(z)$ is equal to some constant, and since it approaches 0 at infinity, is simply 0. Then $f'(z)=A$ (technically this actually leads to a contradiction though, since then $f'(z)$ is bounded, contrary to assumption) for some complex number $A$, and thus $f(z)=Az+B$ for complex numbers $A$ and $B$. To complete the case analysis, if either $f(z)$ or $f'(z)$ were bounded, we could have applied the Liouville's Theorem to see that they were both constants, and thus $f(z)$ would be a polynomial in $z$ of degree at most 1.
Now suppose for the induction hypothesis that $lim_{zrightarrow infty}{frac{f(z)}{z^n}}=0$, we know that $f(z)$ is a polynomial in $z$ of degree at most $n-1$, and consider the case $n+1$. Then if $f(z)$ is not bounded, we know that $$lim_{zrightarrow infty}{frac{f(z)}{z^{n+1}}}=lim_{zrightarrow infty}{frac{f'(z)}{nz^n}}=0$$ so that by our induction hypothesis we know that $f'(z)$ is a polynomial in $z$ of degree at most $n-1$, and thus $f(z)$ is a polynomial in $z$ of degree at most $n$.
Is this proof correct? My main concerns are the use of L'Hospital's Rule, for which there seems to be little information about its validity for complex valued functions. Searching around I could find the following link, but it deals with complex-valued functions of a real variable, rather than the more general complex-valued functions of a complex variable.
complex-analysis analysis analyticity
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
$endgroup$
add a comment |
$begingroup$
I was investigating the following corollary to Liouville's Theorem in Complex Analysis:
if $f(z)$ is entire and $lim_{zrightarrow infty}z^{-n}f(z)=0$, then $f(z)$ is a polynomial in $z$ of degree at most $n-1$.
My proof uses the complex L'Hospital Rule and Induction on $n$. The base case $n=0$ is obvious. To illustrate the idea: for the case $n=2$ Suppose that $f(z)$ is not bounded, so that $$lim_{zrightarrow infty}{f(z)}=infty.$$ Then by using L'Hospital, we have $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}.$$ If $f'(z)$ is not bounded, then we have further $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}=lim_{zrightarrow infty}{frac{f''(z)}{2}}=0.$$ Thus, by Liouville's Theorem, we know that $f''(z)$ is equal to some constant, and since it approaches 0 at infinity, is simply 0. Then $f'(z)=A$ (technically this actually leads to a contradiction though, since then $f'(z)$ is bounded, contrary to assumption) for some complex number $A$, and thus $f(z)=Az+B$ for complex numbers $A$ and $B$. To complete the case analysis, if either $f(z)$ or $f'(z)$ were bounded, we could have applied the Liouville's Theorem to see that they were both constants, and thus $f(z)$ would be a polynomial in $z$ of degree at most 1.
Now suppose for the induction hypothesis that $lim_{zrightarrow infty}{frac{f(z)}{z^n}}=0$, we know that $f(z)$ is a polynomial in $z$ of degree at most $n-1$, and consider the case $n+1$. Then if $f(z)$ is not bounded, we know that $$lim_{zrightarrow infty}{frac{f(z)}{z^{n+1}}}=lim_{zrightarrow infty}{frac{f'(z)}{nz^n}}=0$$ so that by our induction hypothesis we know that $f'(z)$ is a polynomial in $z$ of degree at most $n-1$, and thus $f(z)$ is a polynomial in $z$ of degree at most $n$.
Is this proof correct? My main concerns are the use of L'Hospital's Rule, for which there seems to be little information about its validity for complex valued functions. Searching around I could find the following link, but it deals with complex-valued functions of a real variable, rather than the more general complex-valued functions of a complex variable.
complex-analysis analysis analyticity
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
$endgroup$
1
$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51
add a comment |
$begingroup$
I was investigating the following corollary to Liouville's Theorem in Complex Analysis:
if $f(z)$ is entire and $lim_{zrightarrow infty}z^{-n}f(z)=0$, then $f(z)$ is a polynomial in $z$ of degree at most $n-1$.
My proof uses the complex L'Hospital Rule and Induction on $n$. The base case $n=0$ is obvious. To illustrate the idea: for the case $n=2$ Suppose that $f(z)$ is not bounded, so that $$lim_{zrightarrow infty}{f(z)}=infty.$$ Then by using L'Hospital, we have $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}.$$ If $f'(z)$ is not bounded, then we have further $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}=lim_{zrightarrow infty}{frac{f''(z)}{2}}=0.$$ Thus, by Liouville's Theorem, we know that $f''(z)$ is equal to some constant, and since it approaches 0 at infinity, is simply 0. Then $f'(z)=A$ (technically this actually leads to a contradiction though, since then $f'(z)$ is bounded, contrary to assumption) for some complex number $A$, and thus $f(z)=Az+B$ for complex numbers $A$ and $B$. To complete the case analysis, if either $f(z)$ or $f'(z)$ were bounded, we could have applied the Liouville's Theorem to see that they were both constants, and thus $f(z)$ would be a polynomial in $z$ of degree at most 1.
Now suppose for the induction hypothesis that $lim_{zrightarrow infty}{frac{f(z)}{z^n}}=0$, we know that $f(z)$ is a polynomial in $z$ of degree at most $n-1$, and consider the case $n+1$. Then if $f(z)$ is not bounded, we know that $$lim_{zrightarrow infty}{frac{f(z)}{z^{n+1}}}=lim_{zrightarrow infty}{frac{f'(z)}{nz^n}}=0$$ so that by our induction hypothesis we know that $f'(z)$ is a polynomial in $z$ of degree at most $n-1$, and thus $f(z)$ is a polynomial in $z$ of degree at most $n$.
Is this proof correct? My main concerns are the use of L'Hospital's Rule, for which there seems to be little information about its validity for complex valued functions. Searching around I could find the following link, but it deals with complex-valued functions of a real variable, rather than the more general complex-valued functions of a complex variable.
complex-analysis analysis analyticity
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
$endgroup$
I was investigating the following corollary to Liouville's Theorem in Complex Analysis:
if $f(z)$ is entire and $lim_{zrightarrow infty}z^{-n}f(z)=0$, then $f(z)$ is a polynomial in $z$ of degree at most $n-1$.
My proof uses the complex L'Hospital Rule and Induction on $n$. The base case $n=0$ is obvious. To illustrate the idea: for the case $n=2$ Suppose that $f(z)$ is not bounded, so that $$lim_{zrightarrow infty}{f(z)}=infty.$$ Then by using L'Hospital, we have $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}.$$ If $f'(z)$ is not bounded, then we have further $$lim_{zrightarrow infty}{frac{f(z)}{z^2}}=lim_{zrightarrow infty}{frac{f'(z)}{2z}}=lim_{zrightarrow infty}{frac{f''(z)}{2}}=0.$$ Thus, by Liouville's Theorem, we know that $f''(z)$ is equal to some constant, and since it approaches 0 at infinity, is simply 0. Then $f'(z)=A$ (technically this actually leads to a contradiction though, since then $f'(z)$ is bounded, contrary to assumption) for some complex number $A$, and thus $f(z)=Az+B$ for complex numbers $A$ and $B$. To complete the case analysis, if either $f(z)$ or $f'(z)$ were bounded, we could have applied the Liouville's Theorem to see that they were both constants, and thus $f(z)$ would be a polynomial in $z$ of degree at most 1.
Now suppose for the induction hypothesis that $lim_{zrightarrow infty}{frac{f(z)}{z^n}}=0$, we know that $f(z)$ is a polynomial in $z$ of degree at most $n-1$, and consider the case $n+1$. Then if $f(z)$ is not bounded, we know that $$lim_{zrightarrow infty}{frac{f(z)}{z^{n+1}}}=lim_{zrightarrow infty}{frac{f'(z)}{nz^n}}=0$$ so that by our induction hypothesis we know that $f'(z)$ is a polynomial in $z$ of degree at most $n-1$, and thus $f(z)$ is a polynomial in $z$ of degree at most $n$.
Is this proof correct? My main concerns are the use of L'Hospital's Rule, for which there seems to be little information about its validity for complex valued functions. Searching around I could find the following link, but it deals with complex-valued functions of a real variable, rather than the more general complex-valued functions of a complex variable.
complex-analysis analysis analyticity
complex-analysis analysis analyticity
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
edited Jan 17 at 0:32
Martin Sleziak
44.7k10119272
44.7k10119272
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
asked Mar 7 '14 at 18:14
HaydenHayden
13.7k12347
13.7k12347
1
$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51
add a comment |
1
$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51
1
1
$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This exercise is solved using Cauchy's Integral Formula:
$$
f^{(m)}(0)=frac{m!}{2pi i}int_{|z-z_0|}frac{f(z),dz}{z^{m+1}}
$$
and hence, for $mge n$:
begin{align}
lvert f^{(m)}(0)rvert&=frac{m!}{2pi}left|int_{|z-z_0|=R}frac{f(z),dz}{z^{m+1}}right|
le frac{m!}{2pi}cdot frac{2pi R}{R^{m+1}}max_{|z|=R}lvert f(z)rvert
=m!frac{max_{|z|=R}lvert f(z)rvert}{R^m}\&=
frac{m!}{R^{m-n}}frac{max_{|z|=R}lvert f(z)rvert}{R^n}.
end{align}
By hypothesis
$$
lim_{Rtoinfty}frac{max_{|z|=R}lvert f(z)rvert}{R^n}=0,
$$
and thus $f^{(m)}(0)=0$, for all $mge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n!-!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This exercise is solved using Cauchy's Integral Formula:
$$
f^{(m)}(0)=frac{m!}{2pi i}int_{|z-z_0|}frac{f(z),dz}{z^{m+1}}
$$
and hence, for $mge n$:
begin{align}
lvert f^{(m)}(0)rvert&=frac{m!}{2pi}left|int_{|z-z_0|=R}frac{f(z),dz}{z^{m+1}}right|
le frac{m!}{2pi}cdot frac{2pi R}{R^{m+1}}max_{|z|=R}lvert f(z)rvert
=m!frac{max_{|z|=R}lvert f(z)rvert}{R^m}\&=
frac{m!}{R^{m-n}}frac{max_{|z|=R}lvert f(z)rvert}{R^n}.
end{align}
By hypothesis
$$
lim_{Rtoinfty}frac{max_{|z|=R}lvert f(z)rvert}{R^n}=0,
$$
and thus $f^{(m)}(0)=0$, for all $mge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n!-!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
add a comment |
$begingroup$
This exercise is solved using Cauchy's Integral Formula:
$$
f^{(m)}(0)=frac{m!}{2pi i}int_{|z-z_0|}frac{f(z),dz}{z^{m+1}}
$$
and hence, for $mge n$:
begin{align}
lvert f^{(m)}(0)rvert&=frac{m!}{2pi}left|int_{|z-z_0|=R}frac{f(z),dz}{z^{m+1}}right|
le frac{m!}{2pi}cdot frac{2pi R}{R^{m+1}}max_{|z|=R}lvert f(z)rvert
=m!frac{max_{|z|=R}lvert f(z)rvert}{R^m}\&=
frac{m!}{R^{m-n}}frac{max_{|z|=R}lvert f(z)rvert}{R^n}.
end{align}
By hypothesis
$$
lim_{Rtoinfty}frac{max_{|z|=R}lvert f(z)rvert}{R^n}=0,
$$
and thus $f^{(m)}(0)=0$, for all $mge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n!-!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
add a comment |
$begingroup$
This exercise is solved using Cauchy's Integral Formula:
$$
f^{(m)}(0)=frac{m!}{2pi i}int_{|z-z_0|}frac{f(z),dz}{z^{m+1}}
$$
and hence, for $mge n$:
begin{align}
lvert f^{(m)}(0)rvert&=frac{m!}{2pi}left|int_{|z-z_0|=R}frac{f(z),dz}{z^{m+1}}right|
le frac{m!}{2pi}cdot frac{2pi R}{R^{m+1}}max_{|z|=R}lvert f(z)rvert
=m!frac{max_{|z|=R}lvert f(z)rvert}{R^m}\&=
frac{m!}{R^{m-n}}frac{max_{|z|=R}lvert f(z)rvert}{R^n}.
end{align}
By hypothesis
$$
lim_{Rtoinfty}frac{max_{|z|=R}lvert f(z)rvert}{R^n}=0,
$$
and thus $f^{(m)}(0)=0$, for all $mge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n!-!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
$endgroup$
This exercise is solved using Cauchy's Integral Formula:
$$
f^{(m)}(0)=frac{m!}{2pi i}int_{|z-z_0|}frac{f(z),dz}{z^{m+1}}
$$
and hence, for $mge n$:
begin{align}
lvert f^{(m)}(0)rvert&=frac{m!}{2pi}left|int_{|z-z_0|=R}frac{f(z),dz}{z^{m+1}}right|
le frac{m!}{2pi}cdot frac{2pi R}{R^{m+1}}max_{|z|=R}lvert f(z)rvert
=m!frac{max_{|z|=R}lvert f(z)rvert}{R^m}\&=
frac{m!}{R^{m-n}}frac{max_{|z|=R}lvert f(z)rvert}{R^n}.
end{align}
By hypothesis
$$
lim_{Rtoinfty}frac{max_{|z|=R}lvert f(z)rvert}{R^n}=0,
$$
and thus $f^{(m)}(0)=0$, for all $mge 0$, which means that the power series of $f$ at $z=0$ is a finite sum missing all the terms higher than the $(n!-!1)$-th. Hence $f$ is polynomial of degree at most $n-1$.
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
answered Mar 7 '14 at 18:28
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
63.4k1385163
add a comment |
add a comment |
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$begingroup$
The main problem with your appoach is that even though $f$ is not bounded, you can't conclude that $lim_{ztoinfty} f(z) = infty$. (In fact, that is only true for polynomials. Note for example that $f(z) = e^z$ is not bounded, but $lim_{ztoinfty} e^z$ doesn't exist; not even as $infty$.)
$endgroup$
– mrf
Mar 7 '14 at 18:39
$begingroup$
@mrf, ah, I didn't realize how different the limits at infinity were of complex functions versus real functions.
$endgroup$
– Hayden
Mar 7 '14 at 19:51