If ${u(x,y)+v(x,y): z=x+iyin mathbb{C} }$is bounded then $f$ is constant function.
up vote
0
down vote
favorite
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 at 22:30
ramanujan
651713
add a comment |
up vote
0
down vote
favorite
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 at 22:30
ramanujan
651713
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 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 at 22:42
@John Douma I used this
– ramanujan
Nov 21 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 at 22:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 22:30
ramanujan
651713
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 at 22:30
ramanujan
651713
asked Nov 21 at 22:30
ramanujan
651713
edited Nov 21 at 22:40
asked Nov 21 at 22:30
ramanujan
651713
asked Nov 21 at 22:30
ramanujan
651713
asked Nov 21 at 22:30
ramanujan
651713
651713
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 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 at 22:42
@John Douma I used this
– ramanujan
Nov 21 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 at 22:49
add a comment |
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 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 at 22:42
@John Douma I used this
– ramanujan
Nov 21 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 at 22:49
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 at 22:33
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 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 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 at 22:42
@John Douma I used this
– ramanujan
Nov 21 at 22:45
@John Douma I used this
– ramanujan
Nov 21 at 22:45
1
1
Then I'd say you are good.
– John Douma
Nov 21 at 22:49
Then I'd say you are good.
– John Douma
Nov 21 at 22:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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 at 1:55
Szeto
6,2292726
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 at 1:55
Szeto
6,2292726
add a comment |
up vote
1
down vote
accepted
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 at 1:55
Szeto
6,2292726
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 at 1:55
Szeto
6,2292726
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 at 1:55
Szeto
6,2292726
answered Nov 22 at 1:55
Szeto
6,2292726
answered Nov 22 at 1:55
Szeto
6,2292726
answered Nov 22 at 1:55
Szeto
6,2292726
6,2292726
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008474%2fif-ux-yvx-y-z-xiy-in-mathbbc-is-bounded-then-f-is-constant-fun%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Are you assuming that $f$ is entire?
– Tito Eliatron
Nov 21 at 22:32
@Tito Eliatron yes. I edited it.
– ramanujan
Nov 21 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 at 22:42
@John Douma I used this
– ramanujan
Nov 21 at 22:45
1
Then I'd say you are good.
– John Douma
Nov 21 at 22:49