Prove that if $overline {D_1}subset f(Omega ) $ then $overline {D_r} subset f(Omega )$ for some $r>1$












1












$begingroup$


Let $f:Omega to mathbb C$ be an analytic function such that $Omega $ is an open set.Define $D_r={z:|z|<r}$ for some $r>0$.



Prove that if $overline {D_1}subset f(Omega ) $ then $overline {D_r} subset f(Omega )$ for some $r>1$.



I thought that since $f$ is an analytic function ,it must map an open set into an open set ,so $f(Omega )$ is an open set.



Now $overline {D_1}subset f(Omega ) $ but how can I figure out the $r>1$ for which $overline {D_r} subset f(Omega )$.



This is my first question,I have tried to abide my the rules of this site as far as I could.Hope I will get some help on this problem.






complex-analysis analyticity







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$endgroup$








  • 1




    $begingroup$
    The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
    $endgroup$
    – random123
    Nov 29 '15 at 16:48










  • $begingroup$
    Is it possible to post this as an answer @random123;I did not get your point
    $endgroup$
    – Learnmore
    Nov 29 '15 at 16:55
















1












$begingroup$


Let $f:Omega to mathbb C$ be an analytic function such that $Omega $ is an open set.Define $D_r={z:|z|<r}$ for some $r>0$.



Prove that if $overline {D_1}subset f(Omega ) $ then $overline {D_r} subset f(Omega )$ for some $r>1$.



I thought that since $f$ is an analytic function ,it must map an open set into an open set ,so $f(Omega )$ is an open set.



Now $overline {D_1}subset f(Omega ) $ but how can I figure out the $r>1$ for which $overline {D_r} subset f(Omega )$.



This is my first question,I have tried to abide my the rules of this site as far as I could.Hope I will get some help on this problem.






complex-analysis analyticity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
    $endgroup$
    – random123
    Nov 29 '15 at 16:48










  • $begingroup$
    Is it possible to post this as an answer @random123;I did not get your point
    $endgroup$
    – Learnmore
    Nov 29 '15 at 16:55














1












1








1





$begingroup$


Let $f:Omega to mathbb C$ be an analytic function such that $Omega $ is an open set.Define $D_r={z:|z|<r}$ for some $r>0$.



Prove that if $overline {D_1}subset f(Omega ) $ then $overline {D_r} subset f(Omega )$ for some $r>1$.



I thought that since $f$ is an analytic function ,it must map an open set into an open set ,so $f(Omega )$ is an open set.



Now $overline {D_1}subset f(Omega ) $ but how can I figure out the $r>1$ for which $overline {D_r} subset f(Omega )$.



This is my first question,I have tried to abide my the rules of this site as far as I could.Hope I will get some help on this problem.






complex-analysis analyticity







share|cite|improve this question











$endgroup$




Let $f:Omega to mathbb C$ be an analytic function such that $Omega $ is an open set.Define $D_r={z:|z|<r}$ for some $r>0$.



Prove that if $overline {D_1}subset f(Omega ) $ then $overline {D_r} subset f(Omega )$ for some $r>1$.



I thought that since $f$ is an analytic function ,it must map an open set into an open set ,so $f(Omega )$ is an open set.



Now $overline {D_1}subset f(Omega ) $ but how can I figure out the $r>1$ for which $overline {D_r} subset f(Omega )$.



This is my first question,I have tried to abide my the rules of this site as far as I could.Hope I will get some help on this problem.






complex-analysis analyticity




complex-analysis analyticity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '15 at 14:36







Learnmore

















asked Nov 29 '15 at 14:34









LearnmoreLearnmore

17.7k324100




17.7k324100








  • 1




    $begingroup$
    The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
    $endgroup$
    – random123
    Nov 29 '15 at 16:48










  • $begingroup$
    Is it possible to post this as an answer @random123;I did not get your point
    $endgroup$
    – Learnmore
    Nov 29 '15 at 16:55














  • 1




    $begingroup$
    The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
    $endgroup$
    – random123
    Nov 29 '15 at 16:48










  • $begingroup$
    Is it possible to post this as an answer @random123;I did not get your point
    $endgroup$
    – Learnmore
    Nov 29 '15 at 16:55








1




1




$begingroup$
The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
$endgroup$
– random123
Nov 29 '15 at 16:48




$begingroup$
The minimum distance from $z in overline{D}_1$ and the set $f(Omega)^c$ is a continuous function on $overline{D}_1$ and hence has a minima which will be non-zero(?). Let it be $delta>0$. Then consider $r = 1+ delta$ and show that $D_{r} subset f(Omega)$.
$endgroup$
– random123
Nov 29 '15 at 16:48












$begingroup$
Is it possible to post this as an answer @random123;I did not get your point
$endgroup$
– Learnmore
Nov 29 '15 at 16:55




$begingroup$
Is it possible to post this as an answer @random123;I did not get your point
$endgroup$
– Learnmore
Nov 29 '15 at 16:55










1 Answer
1






active

oldest

votes


















1












$begingroup$

For $z in overline{D}_1$ define a function $alpha(z) = d(z,f(Omega)^c) = inf lbrace d(z,w) | win f(Omega)^c rbrace $ on $overline{D}_1$. Now $alpha$ is a continuous function on $overline{D}_1$(This just follows from triangle inequality). Observe that $f(Omega)^c$ is closed and hence $alpha(z) > 0$(if $alpha(z) = 0$ then exists $w_1 in f(Omega)^c$ such that $d(z,w_i) rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(Omega)^c$ say $w in f(Omega)^c$ since $f(Omega)^c$ is a closed subset of a complete metric space that is $mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w in overline{D}_1 cap f(Omega)^c = emptyset$, a contradiction). Now since $overline{D}_1$ is a compact set, the function $alpha$ has a minima say $delta > 0$ since $alpha$ is a strictly positive function. Let $r = 1 + delta/2$. Then it is easy to see that $overline{D}_r cap f(Omega)^c = emptyset$. Hence $overline{D}_r subset f(Omega)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Shouldn't it be $overline{D}_r subset f(Omega)$?
    $endgroup$
    – Yadati Kiran
    Dec 15 '18 at 5:51










  • $begingroup$
    @YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
    $endgroup$
    – random123
    Dec 15 '18 at 6:02











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$begingroup$

For $z in overline{D}_1$ define a function $alpha(z) = d(z,f(Omega)^c) = inf lbrace d(z,w) | win f(Omega)^c rbrace $ on $overline{D}_1$. Now $alpha$ is a continuous function on $overline{D}_1$(This just follows from triangle inequality). Observe that $f(Omega)^c$ is closed and hence $alpha(z) > 0$(if $alpha(z) = 0$ then exists $w_1 in f(Omega)^c$ such that $d(z,w_i) rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(Omega)^c$ say $w in f(Omega)^c$ since $f(Omega)^c$ is a closed subset of a complete metric space that is $mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w in overline{D}_1 cap f(Omega)^c = emptyset$, a contradiction). Now since $overline{D}_1$ is a compact set, the function $alpha$ has a minima say $delta > 0$ since $alpha$ is a strictly positive function. Let $r = 1 + delta/2$. Then it is easy to see that $overline{D}_r cap f(Omega)^c = emptyset$. Hence $overline{D}_r subset f(Omega)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Shouldn't it be $overline{D}_r subset f(Omega)$?
    $endgroup$
    – Yadati Kiran
    Dec 15 '18 at 5:51










  • $begingroup$
    @YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
    $endgroup$
    – random123
    Dec 15 '18 at 6:02
















1












$begingroup$

For $z in overline{D}_1$ define a function $alpha(z) = d(z,f(Omega)^c) = inf lbrace d(z,w) | win f(Omega)^c rbrace $ on $overline{D}_1$. Now $alpha$ is a continuous function on $overline{D}_1$(This just follows from triangle inequality). Observe that $f(Omega)^c$ is closed and hence $alpha(z) > 0$(if $alpha(z) = 0$ then exists $w_1 in f(Omega)^c$ such that $d(z,w_i) rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(Omega)^c$ say $w in f(Omega)^c$ since $f(Omega)^c$ is a closed subset of a complete metric space that is $mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w in overline{D}_1 cap f(Omega)^c = emptyset$, a contradiction). Now since $overline{D}_1$ is a compact set, the function $alpha$ has a minima say $delta > 0$ since $alpha$ is a strictly positive function. Let $r = 1 + delta/2$. Then it is easy to see that $overline{D}_r cap f(Omega)^c = emptyset$. Hence $overline{D}_r subset f(Omega)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Shouldn't it be $overline{D}_r subset f(Omega)$?
    $endgroup$
    – Yadati Kiran
    Dec 15 '18 at 5:51










  • $begingroup$
    @YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
    $endgroup$
    – random123
    Dec 15 '18 at 6:02














1












1








1





$begingroup$

For $z in overline{D}_1$ define a function $alpha(z) = d(z,f(Omega)^c) = inf lbrace d(z,w) | win f(Omega)^c rbrace $ on $overline{D}_1$. Now $alpha$ is a continuous function on $overline{D}_1$(This just follows from triangle inequality). Observe that $f(Omega)^c$ is closed and hence $alpha(z) > 0$(if $alpha(z) = 0$ then exists $w_1 in f(Omega)^c$ such that $d(z,w_i) rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(Omega)^c$ say $w in f(Omega)^c$ since $f(Omega)^c$ is a closed subset of a complete metric space that is $mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w in overline{D}_1 cap f(Omega)^c = emptyset$, a contradiction). Now since $overline{D}_1$ is a compact set, the function $alpha$ has a minima say $delta > 0$ since $alpha$ is a strictly positive function. Let $r = 1 + delta/2$. Then it is easy to see that $overline{D}_r cap f(Omega)^c = emptyset$. Hence $overline{D}_r subset f(Omega)$.






share|cite|improve this answer











$endgroup$



For $z in overline{D}_1$ define a function $alpha(z) = d(z,f(Omega)^c) = inf lbrace d(z,w) | win f(Omega)^c rbrace $ on $overline{D}_1$. Now $alpha$ is a continuous function on $overline{D}_1$(This just follows from triangle inequality). Observe that $f(Omega)^c$ is closed and hence $alpha(z) > 0$(if $alpha(z) = 0$ then exists $w_1 in f(Omega)^c$ such that $d(z,w_i) rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(Omega)^c$ say $w in f(Omega)^c$ since $f(Omega)^c$ is a closed subset of a complete metric space that is $mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w in overline{D}_1 cap f(Omega)^c = emptyset$, a contradiction). Now since $overline{D}_1$ is a compact set, the function $alpha$ has a minima say $delta > 0$ since $alpha$ is a strictly positive function. Let $r = 1 + delta/2$. Then it is easy to see that $overline{D}_r cap f(Omega)^c = emptyset$. Hence $overline{D}_r subset f(Omega)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 6:03

























answered Nov 29 '15 at 17:22









random123random123

1,2601720




1,2601720








  • 1




    $begingroup$
    Shouldn't it be $overline{D}_r subset f(Omega)$?
    $endgroup$
    – Yadati Kiran
    Dec 15 '18 at 5:51










  • $begingroup$
    @YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
    $endgroup$
    – random123
    Dec 15 '18 at 6:02














  • 1




    $begingroup$
    Shouldn't it be $overline{D}_r subset f(Omega)$?
    $endgroup$
    – Yadati Kiran
    Dec 15 '18 at 5:51










  • $begingroup$
    @YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
    $endgroup$
    – random123
    Dec 15 '18 at 6:02








1




1




$begingroup$
Shouldn't it be $overline{D}_r subset f(Omega)$?
$endgroup$
– Yadati Kiran
Dec 15 '18 at 5:51




$begingroup$
Shouldn't it be $overline{D}_r subset f(Omega)$?
$endgroup$
– Yadati Kiran
Dec 15 '18 at 5:51












$begingroup$
@YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
$endgroup$
– random123
Dec 15 '18 at 6:02




$begingroup$
@YadatiKiran You are right. Thanks for pointing it out. I will make the appropriate edits.
$endgroup$
– random123
Dec 15 '18 at 6:02


















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