Understanding a proof. Why is it necessary that $alpha<1$?
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I'm reading the following proof from Lang's book of complex analysis. But I don't know why is it necessary that $alpha<1$?. I don't know where "$alpha<1$" is used . I don't understand either why $|g_{epsilon}(s)|to 0$ as $|s| to infty$. Can you help me, please?. I have been stuck for many hours.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
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add a comment |
$begingroup$
I'm reading the following proof from Lang's book of complex analysis. But I don't know why is it necessary that $alpha<1$?. I don't know where "$alpha<1$" is used . I don't understand either why $|g_{epsilon}(s)|to 0$ as $|s| to infty$. Can you help me, please?. I have been stuck for many hours.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
$endgroup$
1
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
3
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43
add a comment |
$begingroup$
I'm reading the following proof from Lang's book of complex analysis. But I don't know why is it necessary that $alpha<1$?. I don't know where "$alpha<1$" is used . I don't understand either why $|g_{epsilon}(s)|to 0$ as $|s| to infty$. Can you help me, please?. I have been stuck for many hours.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
$endgroup$
I'm reading the following proof from Lang's book of complex analysis. But I don't know why is it necessary that $alpha<1$?. I don't know where "$alpha<1$" is used . I don't understand either why $|g_{epsilon}(s)|to 0$ as $|s| to infty$. Can you help me, please?. I have been stuck for many hours.
Thank you for any help.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
edited Dec 6 '18 at 22:34
KSM1743
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
asked Dec 6 '18 at 22:25
KSM1743KSM1743
164
164
1
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
3
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43
add a comment |
1
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
3
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43
1
1
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
3
3
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43
add a comment |
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1
$begingroup$
The proof starts with "Let $alpha < beta < 1$". What happens if $beta < alpha$?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:32
$begingroup$
I think that $g_{epsilon}$ might fail to be bounded on the strip. But, if $alpha geq 1$, we could take $beta in (alpha, alpha+1)$ to show the theorem?
$endgroup$
– KSM1743
Dec 6 '18 at 22:39
3
$begingroup$
The proof relies on $0<beta<1$, since that is a requirement for $cos(betasigma) > 0$ for all $sigmain[-pi/2,pi/2]$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 22:43