To show that $|f(z_1)-f(z_2)|leq M|z_1-z_2|$. [closed]
$begingroup$
If $|f'(z)| < M$ for all $z$ on the segment joining complex numbers $z_1$ and $z_2$ then how can I show that $|f(z_1)-f(z_2)|leq M|z_1-z_2|$?
complex-analysis
edited Jan 10 at 5:59
El Pasta
46015
asked Jan 10 at 5:19
user633319user633319
384
$endgroup$
closed as off-topic by Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler Jan 10 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $|f'(z)| < M$ for all $z$ on the segment joining complex numbers $z_1$ and $z_2$ then how can I show that $|f(z_1)-f(z_2)|leq M|z_1-z_2|$?
complex-analysis
edited Jan 10 at 5:59
El Pasta
46015
asked Jan 10 at 5:19
user633319user633319
384
$endgroup$
closed as off-topic by Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler Jan 10 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50
add a comment |
$begingroup$
If $|f'(z)| < M$ for all $z$ on the segment joining complex numbers $z_1$ and $z_2$ then how can I show that $|f(z_1)-f(z_2)|leq M|z_1-z_2|$?
complex-analysis
edited Jan 10 at 5:59
El Pasta
46015
asked Jan 10 at 5:19
user633319user633319
384
$endgroup$
If $|f'(z)| < M$ for all $z$ on the segment joining complex numbers $z_1$ and $z_2$ then how can I show that $|f(z_1)-f(z_2)|leq M|z_1-z_2|$?
complex-analysis
complex-analysis
edited Jan 10 at 5:59
El Pasta
46015
asked Jan 10 at 5:19
user633319user633319
384
edited Jan 10 at 5:59
El Pasta
46015
asked Jan 10 at 5:19
user633319user633319
384
edited Jan 10 at 5:59
El Pasta
46015
edited Jan 10 at 5:59
El Pasta
46015
edited Jan 10 at 5:59
El Pasta
46015
46015
asked Jan 10 at 5:19
user633319user633319
384
asked Jan 10 at 5:19
user633319user633319
384
asked Jan 10 at 5:19
user633319user633319
384
384
closed as off-topic by Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler Jan 10 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler Jan 10 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Lord_Farin, Chickenmancer, Hans Engler
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50
add a comment |
3
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50
3
3
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$(z-z_2)int_0^{1} f'(tz+(1-t)z_2) dt$$ $$=int_0^{1} frac d {dt} f(tz+(1-t)z_2) dt=f(z)-f(z_2)$$ which gives the inequality immediately.
[To see why the first equality holds write down $frac d {dt} f(tz+(1-t)z_2)$ using chain rule].
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
add a comment |
$begingroup$
You can directly think of the line segment as the real line(this is possible since a straight line in the complex plane is homeomorphic to the real line) with the usual order relation(with $z_1>z_2$) and then, by the fundamental theorem of calculus, you'll get
$$|f(z_1)-f(z_2)|=|int_{z_2}^{z_1}f'|leq int_{z_2}^{z_1}|f'|leq int_{z_2}^{z_1}M=M(z_1-z_2)$$
If the order of $z_1$ and $z_2$ were instead $z_2>z_1$, you'll get $|f(z_1)-f(z_2)|leq M(z_2-z_1)$ and hence,
$|f(z_1)-f(z_2)|leq M|z_1-z_2|$
answered Jan 10 at 6:18
MustangMustang
3407
$endgroup$
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$(z-z_2)int_0^{1} f'(tz+(1-t)z_2) dt$$ $$=int_0^{1} frac d {dt} f(tz+(1-t)z_2) dt=f(z)-f(z_2)$$ which gives the inequality immediately.
[To see why the first equality holds write down $frac d {dt} f(tz+(1-t)z_2)$ using chain rule].
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
add a comment |
$begingroup$
$$(z-z_2)int_0^{1} f'(tz+(1-t)z_2) dt$$ $$=int_0^{1} frac d {dt} f(tz+(1-t)z_2) dt=f(z)-f(z_2)$$ which gives the inequality immediately.
[To see why the first equality holds write down $frac d {dt} f(tz+(1-t)z_2)$ using chain rule].
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
add a comment |
$begingroup$
$$(z-z_2)int_0^{1} f'(tz+(1-t)z_2) dt$$ $$=int_0^{1} frac d {dt} f(tz+(1-t)z_2) dt=f(z)-f(z_2)$$ which gives the inequality immediately.
[To see why the first equality holds write down $frac d {dt} f(tz+(1-t)z_2)$ using chain rule].
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
$$(z-z_2)int_0^{1} f'(tz+(1-t)z_2) dt$$ $$=int_0^{1} frac d {dt} f(tz+(1-t)z_2) dt=f(z)-f(z_2)$$ which gives the inequality immediately.
[To see why the first equality holds write down $frac d {dt} f(tz+(1-t)z_2)$ using chain rule].
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
edited Jan 10 at 6:35
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
answered Jan 10 at 5:25
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
54.5k32055
add a comment |
add a comment |
$begingroup$
You can directly think of the line segment as the real line(this is possible since a straight line in the complex plane is homeomorphic to the real line) with the usual order relation(with $z_1>z_2$) and then, by the fundamental theorem of calculus, you'll get
$$|f(z_1)-f(z_2)|=|int_{z_2}^{z_1}f'|leq int_{z_2}^{z_1}|f'|leq int_{z_2}^{z_1}M=M(z_1-z_2)$$
If the order of $z_1$ and $z_2$ were instead $z_2>z_1$, you'll get $|f(z_1)-f(z_2)|leq M(z_2-z_1)$ and hence,
$|f(z_1)-f(z_2)|leq M|z_1-z_2|$
answered Jan 10 at 6:18
MustangMustang
3407
$endgroup$
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
add a comment |
$begingroup$
You can directly think of the line segment as the real line(this is possible since a straight line in the complex plane is homeomorphic to the real line) with the usual order relation(with $z_1>z_2$) and then, by the fundamental theorem of calculus, you'll get
$$|f(z_1)-f(z_2)|=|int_{z_2}^{z_1}f'|leq int_{z_2}^{z_1}|f'|leq int_{z_2}^{z_1}M=M(z_1-z_2)$$
If the order of $z_1$ and $z_2$ were instead $z_2>z_1$, you'll get $|f(z_1)-f(z_2)|leq M(z_2-z_1)$ and hence,
$|f(z_1)-f(z_2)|leq M|z_1-z_2|$
answered Jan 10 at 6:18
MustangMustang
3407
$endgroup$
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
add a comment |
$begingroup$
You can directly think of the line segment as the real line(this is possible since a straight line in the complex plane is homeomorphic to the real line) with the usual order relation(with $z_1>z_2$) and then, by the fundamental theorem of calculus, you'll get
$$|f(z_1)-f(z_2)|=|int_{z_2}^{z_1}f'|leq int_{z_2}^{z_1}|f'|leq int_{z_2}^{z_1}M=M(z_1-z_2)$$
If the order of $z_1$ and $z_2$ were instead $z_2>z_1$, you'll get $|f(z_1)-f(z_2)|leq M(z_2-z_1)$ and hence,
$|f(z_1)-f(z_2)|leq M|z_1-z_2|$
answered Jan 10 at 6:18
MustangMustang
3407
$endgroup$
You can directly think of the line segment as the real line(this is possible since a straight line in the complex plane is homeomorphic to the real line) with the usual order relation(with $z_1>z_2$) and then, by the fundamental theorem of calculus, you'll get
$$|f(z_1)-f(z_2)|=|int_{z_2}^{z_1}f'|leq int_{z_2}^{z_1}|f'|leq int_{z_2}^{z_1}M=M(z_1-z_2)$$
If the order of $z_1$ and $z_2$ were instead $z_2>z_1$, you'll get $|f(z_1)-f(z_2)|leq M(z_2-z_1)$ and hence,
$|f(z_1)-f(z_2)|leq M|z_1-z_2|$
answered Jan 10 at 6:18
MustangMustang
3407
answered Jan 10 at 6:18
MustangMustang
3407
answered Jan 10 at 6:18
MustangMustang
3407
answered Jan 10 at 6:18
MustangMustang
3407
3407
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
add a comment |
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
2
2
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
What do you mean by $z_1 >z_2$? OP is talking about complex numbers.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 6:34
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
$begingroup$
This is because if you imbue the line segment with the subspace topology from the complex plane, the line would become the real line which has its own natural ordering, the only difference is that you would have the option of identifying it in two different ways which would give you either $z_1>z_2$ or $z_2>z_1$(of course $z_1neq z_2$ otherwise the proof is trivial.
$endgroup$
– Mustang
Jan 10 at 6:41
add a comment |
3
$begingroup$
Integrate $f'(z)$ over this segment.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:21
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 10 at 18:50