$f$ analytic and $|f|leq1$ on a strip.
$begingroup$
Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.
The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.
From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.
complex-analysis maximum-principle
edited Dec 21 '18 at 14:36
Song
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
$endgroup$
add a comment |
$begingroup$
Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.
The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.
From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.
complex-analysis maximum-principle
edited Dec 21 '18 at 14:36
Song
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
$endgroup$
$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36
add a comment |
$begingroup$
Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.
The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.
From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.
complex-analysis maximum-principle
edited Dec 21 '18 at 14:36
Song
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
$endgroup$
Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.
The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.
From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.
complex-analysis maximum-principle
complex-analysis maximum-principle
edited Dec 21 '18 at 14:36
Song
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
edited Dec 21 '18 at 14:36
Song
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
edited Dec 21 '18 at 14:36
Song
11.2k628
edited Dec 21 '18 at 14:36
Song
11.2k628
edited Dec 21 '18 at 14:36
Song
11.2k628
11.2k628
asked Dec 20 '18 at 17:32
Ya GYa G
514210
asked Dec 20 '18 at 17:32
Ya GYa G
514210
asked Dec 20 '18 at 17:32
Ya GYa G
514210
514210
$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36
add a comment |
$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36
$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$
answered Dec 20 '18 at 17:47
SongSong
11.2k628
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
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votes
$begingroup$
This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$
answered Dec 20 '18 at 17:47
SongSong
11.2k628
$endgroup$
add a comment |
$begingroup$
This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$
answered Dec 20 '18 at 17:47
SongSong
11.2k628
$endgroup$
add a comment |
$begingroup$
This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$
answered Dec 20 '18 at 17:47
SongSong
11.2k628
$endgroup$
This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$
answered Dec 20 '18 at 17:47
SongSong
11.2k628
answered Dec 20 '18 at 17:47
SongSong
11.2k628
answered Dec 20 '18 at 17:47
SongSong
11.2k628
answered Dec 20 '18 at 17:47
SongSong
11.2k628
11.2k628
add a comment |
add a comment |
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$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58
$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36