Entire function bounded on a set
up vote
0
down vote
favorite
"Consider the set
$T = { alpha ∈ C|∃ a, b ∈ Z :alpha = a + bi }$
Let $g$ be an entire function which satisfies that
$g(z + alpha) = g(z)$
for all $z ∈ mathbb{C}$ and all $alpha ∈ T$.
Prove that $g$ is bounded on the set $ U= {beta ∈ C|∃ x ∈ [0, 1], y ∈
[0, 1] :beta = x + yi}$.
Prove that $g$ is constant on $mathbb{C}$."
I think I have a outline of a proof:
(1) Prove U is compact (closed and bounded)
(2) Use (1) to prove $g$ is bounded on $U$ (Every entire function on a compact subset of $mathbb{C}$ is bounded)
(3) Then, as $g$ is entire by definition and is bounded, I can use Liouville's Theorem to show $g$ is constant (although I feel this only proves $g$ is bounded on $U$ and not on all of $mathbb{C}$)
However, I'm stuck on step (1). I mean, obviously $U$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $g$ is constant everywhere?
general-topology complex-analysis compactness entire-functions
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
asked Nov 20 at 20:37
Dino
836
add a comment |
up vote
0
down vote
favorite
"Consider the set
$T = { alpha ∈ C|∃ a, b ∈ Z :alpha = a + bi }$
Let $g$ be an entire function which satisfies that
$g(z + alpha) = g(z)$
for all $z ∈ mathbb{C}$ and all $alpha ∈ T$.
Prove that $g$ is bounded on the set $ U= {beta ∈ C|∃ x ∈ [0, 1], y ∈
[0, 1] :beta = x + yi}$.
Prove that $g$ is constant on $mathbb{C}$."
I think I have a outline of a proof:
(1) Prove U is compact (closed and bounded)
(2) Use (1) to prove $g$ is bounded on $U$ (Every entire function on a compact subset of $mathbb{C}$ is bounded)
(3) Then, as $g$ is entire by definition and is bounded, I can use Liouville's Theorem to show $g$ is constant (although I feel this only proves $g$ is bounded on $U$ and not on all of $mathbb{C}$)
However, I'm stuck on step (1). I mean, obviously $U$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $g$ is constant everywhere?
general-topology complex-analysis compactness entire-functions
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
asked Nov 20 at 20:37
Dino
836
(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
"Consider the set
$T = { alpha ∈ C|∃ a, b ∈ Z :alpha = a + bi }$
Let $g$ be an entire function which satisfies that
$g(z + alpha) = g(z)$
for all $z ∈ mathbb{C}$ and all $alpha ∈ T$.
Prove that $g$ is bounded on the set $ U= {beta ∈ C|∃ x ∈ [0, 1], y ∈
[0, 1] :beta = x + yi}$.
Prove that $g$ is constant on $mathbb{C}$."
I think I have a outline of a proof:
(1) Prove U is compact (closed and bounded)
(2) Use (1) to prove $g$ is bounded on $U$ (Every entire function on a compact subset of $mathbb{C}$ is bounded)
(3) Then, as $g$ is entire by definition and is bounded, I can use Liouville's Theorem to show $g$ is constant (although I feel this only proves $g$ is bounded on $U$ and not on all of $mathbb{C}$)
However, I'm stuck on step (1). I mean, obviously $U$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $g$ is constant everywhere?
general-topology complex-analysis compactness entire-functions
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
asked Nov 20 at 20:37
Dino
836
"Consider the set
$T = { alpha ∈ C|∃ a, b ∈ Z :alpha = a + bi }$
Let $g$ be an entire function which satisfies that
$g(z + alpha) = g(z)$
for all $z ∈ mathbb{C}$ and all $alpha ∈ T$.
Prove that $g$ is bounded on the set $ U= {beta ∈ C|∃ x ∈ [0, 1], y ∈
[0, 1] :beta = x + yi}$.
Prove that $g$ is constant on $mathbb{C}$."
I think I have a outline of a proof:
(1) Prove U is compact (closed and bounded)
(2) Use (1) to prove $g$ is bounded on $U$ (Every entire function on a compact subset of $mathbb{C}$ is bounded)
(3) Then, as $g$ is entire by definition and is bounded, I can use Liouville's Theorem to show $g$ is constant (although I feel this only proves $g$ is bounded on $U$ and not on all of $mathbb{C}$)
However, I'm stuck on step (1). I mean, obviously $U$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $g$ is constant everywhere?
general-topology complex-analysis compactness entire-functions
general-topology complex-analysis compactness entire-functions
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
asked Nov 20 at 20:37
Dino
836
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
asked Nov 20 at 20:37
Dino
836
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
edited Nov 20 at 20:59
José Carlos Santos
147k22117217
147k22117217
asked Nov 20 at 20:37
Dino
836
asked Nov 20 at 20:37
Dino
836
asked Nov 20 at 20:37
Dino
836
836
(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41
add a comment |
(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41
(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41
(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{ninmathbb N}$ of elements of $U$ converges to $zinmathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $zin U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $ain T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{ninmathbb N}$ of elements of $U$ converges to $zinmathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $zin U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $ain T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
add a comment |
up vote
2
down vote
accepted
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{ninmathbb N}$ of elements of $U$ converges to $zinmathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $zin U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $ain T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{ninmathbb N}$ of elements of $U$ converges to $zinmathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $zin U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $ain T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{ninmathbb N}$ of elements of $U$ converges to $zinmathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $zin U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $ain T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
answered Nov 20 at 20:58
José Carlos Santos
147k22117217
147k22117217
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
add a comment |
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded
– Dino
Nov 21 at 17:32
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
Let $z$ be any complex number. Then$$z=z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor i+lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor i$$and therefore$$g(z)=gbigl(z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor ibigr),$$since $lflooroperatorname{Re}zrfloor+lflooroperatorname{Im}zrfloor iin T$. But $z-lflooroperatorname{Re}zrfloor-lflooroperatorname{Im}zrfloor iin U$ and so $bigllvert g(z)bigrrvertleqslantsuplvert grvert(U)$.
– José Carlos Santos
Nov 21 at 18:12
add a comment |
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(1) Use the Heine-Borel theorem
– Lord Shark the Unknown
Nov 20 at 20:41