Mixing
If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is bounded then $f$ is constant function.
Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.
Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you
complex-analysis
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
add a comment |
Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.
Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you
complex-analysis
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49
add a comment |
Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.
Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you
complex-analysis
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is
bounded then function is constant.
Here is what I tried.
Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you
complex-analysis
complex-analysis
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
edited Nov 21 '18 at 22:40
ramanujan
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
asked Nov 21 '18 at 22:30
ramanujanramanujan
714713
714713
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49
add a comment |
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
1
1
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49
add a comment |
1 Answer
1
active
oldest
votes
Another approach:
We have the identity
$$leftvert e^z right vert=e^{Re z}$$
Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.
We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$
Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$
which completes the proof.
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Another approach:
We have the identity
$$leftvert e^z right vert=e^{Re z}$$
Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.
We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$
Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$
which completes the proof.
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
add a comment |
Another approach:
We have the identity
$$leftvert e^z right vert=e^{Re z}$$
Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.
We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$
Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$
which completes the proof.
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
add a comment |
Another approach:
We have the identity
$$leftvert e^z right vert=e^{Re z}$$
Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.
We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$
Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$
which completes the proof.
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
Another approach:
We have the identity
$$leftvert e^z right vert=e^{Re z}$$
Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.
We have
$$|exp g(z)|=leftvert e^{(1-i)f(z)}rightvert=e^{u+v}le e^M$$
Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=Cimplies g(z)=text{constant}implies f(z)=text{constant}$$
which completes the proof.
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
answered Nov 22 '18 at 1:55
SzetoSzeto
6,4362926
6,4362926
add a comment |
add a comment |
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Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 '18 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 '18 at 22:33
This looks good but you need that both the real and imaginary parts of $g$ are bounded so you can say all of $g$ is bounded. Then it follows that $g$ is constant.
– John Douma
Nov 21 '18 at 22:42
@John Douma I used this
– ramanujan
Nov 21 '18 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 '18 at 22:49