Mixing
Analytic contiunation
0
$begingroup$
this is more of a broader question.
Say I have an analytic complex function $f$ that is defined on the open unit circle, but I know that the limit of $f$ when $|z|$ approaches 1 is 0. Can you define a function $g$ that satisfies:
for $|z| lt 1$, $g(z) = f(z)$
for $|z| = 1$ $g(z) = 0$
And would that function remain analytic?
complex-analysis analytic-functions analytic-continuation
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
add a comment |
0
$begingroup$
this is more of a broader question.
Say I have an analytic complex function $f$ that is defined on the open unit circle, but I know that the limit of $f$ when $|z|$ approaches 1 is 0. Can you define a function $g$ that satisfies:
for $|z| lt 1$, $g(z) = f(z)$
for $|z| = 1$ $g(z) = 0$
And would that function remain analytic?
complex-analysis analytic-functions analytic-continuation
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16
add a comment |
0
0
0
$begingroup$
this is more of a broader question.
Say I have an analytic complex function $f$ that is defined on the open unit circle, but I know that the limit of $f$ when $|z|$ approaches 1 is 0. Can you define a function $g$ that satisfies:
for $|z| lt 1$, $g(z) = f(z)$
for $|z| = 1$ $g(z) = 0$
And would that function remain analytic?
complex-analysis analytic-functions analytic-continuation
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
$endgroup$
this is more of a broader question.
Say I have an analytic complex function $f$ that is defined on the open unit circle, but I know that the limit of $f$ when $|z|$ approaches 1 is 0. Can you define a function $g$ that satisfies:
for $|z| lt 1$, $g(z) = f(z)$
for $|z| = 1$ $g(z) = 0$
And would that function remain analytic?
complex-analysis analytic-functions analytic-continuation
complex-analysis analytic-functions analytic-continuation
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
asked Jan 26 at 14:39
Guy SchwartzbergGuy Schwartzberg
1247
1247
$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16
add a comment |
$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16
$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16
add a comment |
0
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$begingroup$
Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic?
$endgroup$
– Yanko
Jan 26 at 14:46
$begingroup$
Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| le Re(z)$, prove $f = 0$
$endgroup$
– Guy Schwartzberg
Jan 26 at 15:03
$begingroup$
Don't your hypotheses imply that $f$ is identically zero in the open disk?
$endgroup$
– Andreas Blass
Jan 26 at 18:19
$begingroup$
Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem
$endgroup$
– Guy Schwartzberg
Jan 27 at 10:16