Let $I⊂mathbb R$ be an interval and suppose $f:I→mathbb R$ is differentiable and $f':I→mathbb R$ is...
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Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. For $L>0$, a function $f : X → Y$ is $L$-Lipschitz continuous if$$d_Y(f(x), f(x')) ≤ Ld_X(x,x').quadforall x, x'in X$$We say $f:X→Y$ is Lipschitz continuous if $f$ is $L$-Lipschitz continuous for some $L>0$.
(a) Show that if $f : X →Y$ is Lipschitz continuous, then $f$ is continuous.
(b) Let $I ⊂ mathbb R$ be an interval and suppose $f : I → mathbb R$ is differentiable and $f': I → mathbb R$ is bounded. Show $f$ is Lipschitz continuous.
*For part a, is showing $f$ is uniformly continuous okay? or I can only show it is continuous?
*For part b, I do not know how to do this at all.
real-analysis
edited Nov 30 '18 at 10:08
Saad
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
$endgroup$
add a comment |
$begingroup$
Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. For $L>0$, a function $f : X → Y$ is $L$-Lipschitz continuous if$$d_Y(f(x), f(x')) ≤ Ld_X(x,x').quadforall x, x'in X$$We say $f:X→Y$ is Lipschitz continuous if $f$ is $L$-Lipschitz continuous for some $L>0$.
(a) Show that if $f : X →Y$ is Lipschitz continuous, then $f$ is continuous.
(b) Let $I ⊂ mathbb R$ be an interval and suppose $f : I → mathbb R$ is differentiable and $f': I → mathbb R$ is bounded. Show $f$ is Lipschitz continuous.
*For part a, is showing $f$ is uniformly continuous okay? or I can only show it is continuous?
*For part b, I do not know how to do this at all.
real-analysis
edited Nov 30 '18 at 10:08
Saad
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
$endgroup$
$begingroup$
This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47
add a comment |
$begingroup$
Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. For $L>0$, a function $f : X → Y$ is $L$-Lipschitz continuous if$$d_Y(f(x), f(x')) ≤ Ld_X(x,x').quadforall x, x'in X$$We say $f:X→Y$ is Lipschitz continuous if $f$ is $L$-Lipschitz continuous for some $L>0$.
(a) Show that if $f : X →Y$ is Lipschitz continuous, then $f$ is continuous.
(b) Let $I ⊂ mathbb R$ be an interval and suppose $f : I → mathbb R$ is differentiable and $f': I → mathbb R$ is bounded. Show $f$ is Lipschitz continuous.
*For part a, is showing $f$ is uniformly continuous okay? or I can only show it is continuous?
*For part b, I do not know how to do this at all.
real-analysis
edited Nov 30 '18 at 10:08
Saad
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
$endgroup$
Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. For $L>0$, a function $f : X → Y$ is $L$-Lipschitz continuous if$$d_Y(f(x), f(x')) ≤ Ld_X(x,x').quadforall x, x'in X$$We say $f:X→Y$ is Lipschitz continuous if $f$ is $L$-Lipschitz continuous for some $L>0$.
(a) Show that if $f : X →Y$ is Lipschitz continuous, then $f$ is continuous.
(b) Let $I ⊂ mathbb R$ be an interval and suppose $f : I → mathbb R$ is differentiable and $f': I → mathbb R$ is bounded. Show $f$ is Lipschitz continuous.
*For part a, is showing $f$ is uniformly continuous okay? or I can only show it is continuous?
*For part b, I do not know how to do this at all.
real-analysis
real-analysis
edited Nov 30 '18 at 10:08
Saad
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
edited Nov 30 '18 at 10:08
Saad
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
edited Nov 30 '18 at 10:08
Saad
19.7k92352
edited Nov 30 '18 at 10:08
Saad
19.7k92352
edited Nov 30 '18 at 10:08
Saad
19.7k92352
19.7k92352
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
asked Nov 30 '18 at 7:41
Math AvengersMath Avengers
709
709
$begingroup$
This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47
add a comment |
$begingroup$
This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47
$begingroup$
This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47
$begingroup$
This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47
add a comment |
1 Answer
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Uniformly continuous functions are continuous
For b) use Mean Value Theorem: $|f(x)-f(y)| =|x-y| |f'(xi)|$ for some $xi$ which gives $|f(x)-f(y)| leq L|x-y|$ where $L$ is an upper bound for $|f'|$ .
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
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1 Answer
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1 Answer
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active
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votes
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active
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$begingroup$
Uniformly continuous functions are continuous
For b) use Mean Value Theorem: $|f(x)-f(y)| =|x-y| |f'(xi)|$ for some $xi$ which gives $|f(x)-f(y)| leq L|x-y|$ where $L$ is an upper bound for $|f'|$ .
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
add a comment |
$begingroup$
Uniformly continuous functions are continuous
For b) use Mean Value Theorem: $|f(x)-f(y)| =|x-y| |f'(xi)|$ for some $xi$ which gives $|f(x)-f(y)| leq L|x-y|$ where $L$ is an upper bound for $|f'|$ .
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
add a comment |
$begingroup$
Uniformly continuous functions are continuous
For b) use Mean Value Theorem: $|f(x)-f(y)| =|x-y| |f'(xi)|$ for some $xi$ which gives $|f(x)-f(y)| leq L|x-y|$ where $L$ is an upper bound for $|f'|$ .
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
Uniformly continuous functions are continuous
For b) use Mean Value Theorem: $|f(x)-f(y)| =|x-y| |f'(xi)|$ for some $xi$ which gives $|f(x)-f(y)| leq L|x-y|$ where $L$ is an upper bound for $|f'|$ .
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Nov 30 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
52.6k32055
add a comment |
add a comment |
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This is just an immediate consequence of the mean value theorem. Showing uniform continuity is fine, since uniform continuity is stronger than continuity
$endgroup$
– MathematicsStudent1122
Nov 30 '18 at 7:47