Mixing
Should I use Liouville thm or continuity?
$begingroup$
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the properties:(Justify your answer)
$|f(z)|leq frac{1}{100}$ if $|z|leq 1$ and
$|f(z)-1|leq frac{1}{100}$ if $|z-3|leq 1$.
My attemp: I confuesd to use Liouville thm or continuity?
Please guide me how should I think about it!
complex-analysis
asked Jan 9 at 19:48
seyedseyed
566
$endgroup$
add a comment |
$begingroup$
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the properties:(Justify your answer)
$|f(z)|leq frac{1}{100}$ if $|z|leq 1$ and
$|f(z)-1|leq frac{1}{100}$ if $|z-3|leq 1$.
My attemp: I confuesd to use Liouville thm or continuity?
Please guide me how should I think about it!
complex-analysis
asked Jan 9 at 19:48
seyedseyed
566
$endgroup$
$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57
add a comment |
$begingroup$
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the properties:(Justify your answer)
$|f(z)|leq frac{1}{100}$ if $|z|leq 1$ and
$|f(z)-1|leq frac{1}{100}$ if $|z-3|leq 1$.
My attemp: I confuesd to use Liouville thm or continuity?
Please guide me how should I think about it!
complex-analysis
asked Jan 9 at 19:48
seyedseyed
566
$endgroup$
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the properties:(Justify your answer)
$|f(z)|leq frac{1}{100}$ if $|z|leq 1$ and
$|f(z)-1|leq frac{1}{100}$ if $|z-3|leq 1$.
My attemp: I confuesd to use Liouville thm or continuity?
Please guide me how should I think about it!
complex-analysis
complex-analysis
asked Jan 9 at 19:48
seyedseyed
566
asked Jan 9 at 19:48
seyedseyed
566
asked Jan 9 at 19:48
seyedseyed
566
asked Jan 9 at 19:48
seyedseyed
566
asked Jan 9 at 19:48
seyedseyed
566
566
$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57
add a comment |
$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57
$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57
$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let us define $g(z)=0$ on $|z|le 1$ and $g(z)=1$ on $|z-3|le 1$. Then $g$ is analytic on a neigoborhood of $K={|z|le 1}cup {|z-3|le 1}$. Note that $Bbb Csetminus K$ is connected. By Runge's theorem, for every $epsilon>0$, there exists a polynomial $p(z)$ such that
$$
|g(z)-p(z)|leepsilon,quadforall zin K.
$$ If we let $epsilon =frac{1}{100}$, then $p$ is a function we are looking for.
answered Jan 15 at 10:16
SongSong
11.5k628
$endgroup$
add a comment |
$begingroup$
You want an entire function satisfying the given two properties.
What about the constant function $f(z)=1/200$ ?
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
$endgroup$
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us define $g(z)=0$ on $|z|le 1$ and $g(z)=1$ on $|z-3|le 1$. Then $g$ is analytic on a neigoborhood of $K={|z|le 1}cup {|z-3|le 1}$. Note that $Bbb Csetminus K$ is connected. By Runge's theorem, for every $epsilon>0$, there exists a polynomial $p(z)$ such that
$$
|g(z)-p(z)|leepsilon,quadforall zin K.
$$ If we let $epsilon =frac{1}{100}$, then $p$ is a function we are looking for.
answered Jan 15 at 10:16
SongSong
11.5k628
$endgroup$
add a comment |
$begingroup$
Let us define $g(z)=0$ on $|z|le 1$ and $g(z)=1$ on $|z-3|le 1$. Then $g$ is analytic on a neigoborhood of $K={|z|le 1}cup {|z-3|le 1}$. Note that $Bbb Csetminus K$ is connected. By Runge's theorem, for every $epsilon>0$, there exists a polynomial $p(z)$ such that
$$
|g(z)-p(z)|leepsilon,quadforall zin K.
$$ If we let $epsilon =frac{1}{100}$, then $p$ is a function we are looking for.
answered Jan 15 at 10:16
SongSong
11.5k628
$endgroup$
add a comment |
$begingroup$
Let us define $g(z)=0$ on $|z|le 1$ and $g(z)=1$ on $|z-3|le 1$. Then $g$ is analytic on a neigoborhood of $K={|z|le 1}cup {|z-3|le 1}$. Note that $Bbb Csetminus K$ is connected. By Runge's theorem, for every $epsilon>0$, there exists a polynomial $p(z)$ such that
$$
|g(z)-p(z)|leepsilon,quadforall zin K.
$$ If we let $epsilon =frac{1}{100}$, then $p$ is a function we are looking for.
answered Jan 15 at 10:16
SongSong
11.5k628
$endgroup$
Let us define $g(z)=0$ on $|z|le 1$ and $g(z)=1$ on $|z-3|le 1$. Then $g$ is analytic on a neigoborhood of $K={|z|le 1}cup {|z-3|le 1}$. Note that $Bbb Csetminus K$ is connected. By Runge's theorem, for every $epsilon>0$, there exists a polynomial $p(z)$ such that
$$
|g(z)-p(z)|leepsilon,quadforall zin K.
$$ If we let $epsilon =frac{1}{100}$, then $p$ is a function we are looking for.
answered Jan 15 at 10:16
SongSong
11.5k628
answered Jan 15 at 10:16
SongSong
11.5k628
answered Jan 15 at 10:16
SongSong
11.5k628
answered Jan 15 at 10:16
SongSong
11.5k628
11.5k628
add a comment |
add a comment |
$begingroup$
You want an entire function satisfying the given two properties.
What about the constant function $f(z)=1/200$ ?
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
$endgroup$
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
add a comment |
$begingroup$
You want an entire function satisfying the given two properties.
What about the constant function $f(z)=1/200$ ?
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
$endgroup$
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
add a comment |
$begingroup$
You want an entire function satisfying the given two properties.
What about the constant function $f(z)=1/200$ ?
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
$endgroup$
You want an entire function satisfying the given two properties.
What about the constant function $f(z)=1/200$ ?
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
answered Jan 10 at 7:13
EmptyEmpty
8,12652661
8,12652661
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
add a comment |
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
$begingroup$
$|1/200 - 1| = 199/200 > 1/100$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 10 at 10:31
add a comment |
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$begingroup$
If $g$ is an entire function, it is continuous in particular, and hence, bounded by some constant $a = a(D) > 0$ on the disc $D.$
$endgroup$
– Will M.
Jan 9 at 19:57