Problem: Is function unifomly continuous?
Is sin ( 1/x ) uniformly continuous on set (1, + infinity) ?
I tried to prove that it is Lipschitz continuous but I got stuck.
uniform-continuity
asked Nov 18 at 15:42
user560461
525
add a comment |
Is sin ( 1/x ) uniformly continuous on set (1, + infinity) ?
I tried to prove that it is Lipschitz continuous but I got stuck.
uniform-continuity
asked Nov 18 at 15:42
user560461
525
add a comment |
Is sin ( 1/x ) uniformly continuous on set (1, + infinity) ?
I tried to prove that it is Lipschitz continuous but I got stuck.
uniform-continuity
asked Nov 18 at 15:42
user560461
525
Is sin ( 1/x ) uniformly continuous on set (1, + infinity) ?
I tried to prove that it is Lipschitz continuous but I got stuck.
uniform-continuity
uniform-continuity
asked Nov 18 at 15:42
user560461
525
asked Nov 18 at 15:42
user560461
525
asked Nov 18 at 15:42
user560461
525
asked Nov 18 at 15:42
user560461
525
asked Nov 18 at 15:42
user560461
525
525
add a comment |
add a comment |
1 Answer
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Certainly: it's differentiable with bounded first derivative. First property of Lipschitz continuous functions as listed on Wikipedia:
An everywhere differentiable function $g : mathbb{R} to mathbb{R}$ is Lipschitz continuous (with $K = sup |g'(x)|$) if and only if it has bounded first derivative; one direction follows from the mean value theorem."
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
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1 Answer
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1 Answer
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Certainly: it's differentiable with bounded first derivative. First property of Lipschitz continuous functions as listed on Wikipedia:
An everywhere differentiable function $g : mathbb{R} to mathbb{R}$ is Lipschitz continuous (with $K = sup |g'(x)|$) if and only if it has bounded first derivative; one direction follows from the mean value theorem."
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
add a comment |
Certainly: it's differentiable with bounded first derivative. First property of Lipschitz continuous functions as listed on Wikipedia:
An everywhere differentiable function $g : mathbb{R} to mathbb{R}$ is Lipschitz continuous (with $K = sup |g'(x)|$) if and only if it has bounded first derivative; one direction follows from the mean value theorem."
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
add a comment |
Certainly: it's differentiable with bounded first derivative. First property of Lipschitz continuous functions as listed on Wikipedia:
An everywhere differentiable function $g : mathbb{R} to mathbb{R}$ is Lipschitz continuous (with $K = sup |g'(x)|$) if and only if it has bounded first derivative; one direction follows from the mean value theorem."
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
Certainly: it's differentiable with bounded first derivative. First property of Lipschitz continuous functions as listed on Wikipedia:
An everywhere differentiable function $g : mathbb{R} to mathbb{R}$ is Lipschitz continuous (with $K = sup |g'(x)|$) if and only if it has bounded first derivative; one direction follows from the mean value theorem."
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
answered Nov 18 at 15:48
Patrick Stevens
28.5k52874
28.5k52874
add a comment |
add a comment |
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