Analytic and bounded implies uniform continuity
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Let $f$ be analytic and bounded in $zinmathbbCmid Re(z)>0$. Prove that $f$ is uniformly continuous in $zinmathbbCmid Re(z)>C=:D$ for every $C>0.$
For uniform continuity, I have to show that for every $varepsilon>0$ there exists $delta>0$ such that for all $x,yin D$ with $|x-y|<delta$ we have $|f(x)-f(y)|<varepsilon$. How can I find such a $delta$? I don't know how to use the assumptions that $f$ is analytic (i.e. can be written as a power series) and bounded.
complex-analysis uniform-convergence analytic-functions
edited Aug 16 at 23:35
zhw.
66.6k42871
asked Aug 16 at 22:59
mathstackuser
64911
add a comment |Â
up vote
3
down vote
favorite
Let $f$ be analytic and bounded in $zinmathbbCmid Re(z)>0$. Prove that $f$ is uniformly continuous in $zinmathbbCmid Re(z)>C=:D$ for every $C>0.$
For uniform continuity, I have to show that for every $varepsilon>0$ there exists $delta>0$ such that for all $x,yin D$ with $|x-y|<delta$ we have $|f(x)-f(y)|<varepsilon$. How can I find such a $delta$? I don't know how to use the assumptions that $f$ is analytic (i.e. can be written as a power series) and bounded.
complex-analysis uniform-convergence analytic-functions
edited Aug 16 at 23:35
zhw.
66.6k42871
asked Aug 16 at 22:59
mathstackuser
64911
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f$ be analytic and bounded in $zinmathbbCmid Re(z)>0$. Prove that $f$ is uniformly continuous in $zinmathbbCmid Re(z)>C=:D$ for every $C>0.$
For uniform continuity, I have to show that for every $varepsilon>0$ there exists $delta>0$ such that for all $x,yin D$ with $|x-y|<delta$ we have $|f(x)-f(y)|<varepsilon$. How can I find such a $delta$? I don't know how to use the assumptions that $f$ is analytic (i.e. can be written as a power series) and bounded.
complex-analysis uniform-convergence analytic-functions
edited Aug 16 at 23:35
zhw.
66.6k42871
asked Aug 16 at 22:59
mathstackuser
64911
Let $f$ be analytic and bounded in $zinmathbbCmid Re(z)>0$. Prove that $f$ is uniformly continuous in $zinmathbbCmid Re(z)>C=:D$ for every $C>0.$
For uniform continuity, I have to show that for every $varepsilon>0$ there exists $delta>0$ such that for all $x,yin D$ with $|x-y|<delta$ we have $|f(x)-f(y)|<varepsilon$. How can I find such a $delta$? I don't know how to use the assumptions that $f$ is analytic (i.e. can be written as a power series) and bounded.
complex-analysis uniform-convergence analytic-functions
edited Aug 16 at 23:35
zhw.
66.6k42871
asked Aug 16 at 22:59
mathstackuser
64911
edited Aug 16 at 23:35
zhw.
66.6k42871
edited Aug 16 at 23:35
zhw.
66.6k42871
edited Aug 16 at 23:35
zhw.
66.6k42871
66.6k42871
asked Aug 16 at 22:59
mathstackuser
64911
asked Aug 16 at 22:59
mathstackuser
64911
asked Aug 16 at 22:59
mathstackuser
64911
64911
add a comment |Â
add a comment |Â
1 Answer
1
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votes
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2
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Hint: Suppose $|f|le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|le M/x$ everywhere in the right half plane.
answered Aug 16 at 23:36
zhw.
66.6k42871
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: Suppose $|f|le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|le M/x$ everywhere in the right half plane.
answered Aug 16 at 23:36
zhw.
66.6k42871
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
add a comment |Â
up vote
2
down vote
Hint: Suppose $|f|le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|le M/x$ everywhere in the right half plane.
answered Aug 16 at 23:36
zhw.
66.6k42871
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Suppose $|f|le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|le M/x$ everywhere in the right half plane.
answered Aug 16 at 23:36
zhw.
66.6k42871
Hint: Suppose $|f|le M$ in the right half plane. Cauchy's estimates give $|f'(x+iy)|le M/x$ everywhere in the right half plane.
answered Aug 16 at 23:36
zhw.
66.6k42871
answered Aug 16 at 23:36
zhw.
66.6k42871
answered Aug 16 at 23:36
zhw.
66.6k42871
answered Aug 16 at 23:36
zhw.
66.6k42871
66.6k42871
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
add a comment |Â
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
Thanks! But how can I use the derivative to show uniform continuity?
– mathstackuser
Aug 17 at 11:29
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
@mathstackuser What would you do if you had a function on the real line whose derivative was $le 1$ in absolute value everywhere?
– zhw.
Aug 17 at 17:44
add a comment |Â
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