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Can one deduce the uniformly continuity by a “weak” Lipschtz condition?

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1

$begingroup$

$exists L>0,$ such that for an arbitrary subinterval $Isubset[a,infty),exists x_1,x_2in I$,we have $$|f(x_1)-f(x_2)|<L|x_1-x_2|$$

This condition is much weaker than the Lipschtiz condition , can one deduce the the uniformly continuity of $f$ on $[a,infty)$ ?

calculus

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asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

$endgroup$

add a comment | 

1

$begingroup$

$exists L>0,$ such that for an arbitrary subinterval $Isubset[a,infty),exists x_1,x_2in I$,we have $$|f(x_1)-f(x_2)|<L|x_1-x_2|$$

This condition is much weaker than the Lipschtiz condition , can one deduce the the uniformly continuity of $f$ on $[a,infty)$ ?

calculus

share|cite|improve this question

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

$endgroup$

add a comment | 

$begingroup$

$exists L>0,$ such that for an arbitrary subinterval $Isubset[a,infty),exists x_1,x_2in I$,we have $$|f(x_1)-f(x_2)|<L|x_1-x_2|$$

This condition is much weaker than the Lipschtiz condition , can one deduce the the uniformly continuity of $f$ on $[a,infty)$ ?

calculus

share|cite|improve this question

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

$endgroup$

share|cite|improve this question

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

share|cite|improve this question

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

asked Mar 18 at 13:36

Alexander LauAlexander Lau

1348

2 Answers
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No, one cannot. In fact, there are discontinuous functions satisfying the condition. Take for example the famous function continuous only at the irrationals: $f(x)=0$ if $x$ is irrational, and $f(p/q)=1/q$ if $p/q$ is a rational number given in its lowest terms. It can be shown that $f$ is continuous precisely at the irrational points. It obviously satisfies your condition with $L=1$ because every interval contains distinct irrational numbers. However, it cannot be uniformly continuous, because it is not continuous.

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answered Mar 18 at 13:47

uniquesolutionuniquesolution

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1

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The non-uniformly continuous function, $C^1$ function $f(x) = x sin x$ satisfies your criterion, because it has a sequence of critical points $x_n$ that approach infinity. For any critical point $x$ of any $C^1$ function $f$ and for any value of $L$, there exists $delta>0$ such that your inequality is true for all $x_1,x_2 in (x-delta,x+delta)$.

When I think of the $L$-Lipschitz condition, I like to remember one of its properties: if $f$ is differentiable and $L$-Lipschitz then $|f'(x)| le L$ for all $x$ in the domain of $f$. I concocted my example so that $f(x)$ has two phenomena: many critical points, and points with larger and larger derivative.

share|cite|improve this answer

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

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1

$begingroup$

No, one cannot. In fact, there are discontinuous functions satisfying the condition. Take for example the famous function continuous only at the irrationals: $f(x)=0$ if $x$ is irrational, and $f(p/q)=1/q$ if $p/q$ is a rational number given in its lowest terms. It can be shown that $f$ is continuous precisely at the irrational points. It obviously satisfies your condition with $L=1$ because every interval contains distinct irrational numbers. However, it cannot be uniformly continuous, because it is not continuous.

share|cite|improve this answer

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

$endgroup$

add a comment | 

1

$begingroup$

No, one cannot. In fact, there are discontinuous functions satisfying the condition. Take for example the famous function continuous only at the irrationals: $f(x)=0$ if $x$ is irrational, and $f(p/q)=1/q$ if $p/q$ is a rational number given in its lowest terms. It can be shown that $f$ is continuous precisely at the irrational points. It obviously satisfies your condition with $L=1$ because every interval contains distinct irrational numbers. However, it cannot be uniformly continuous, because it is not continuous.

share|cite|improve this answer

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

$endgroup$

add a comment | 

$begingroup$

No, one cannot. In fact, there are discontinuous functions satisfying the condition. Take for example the famous function continuous only at the irrationals: $f(x)=0$ if $x$ is irrational, and $f(p/q)=1/q$ if $p/q$ is a rational number given in its lowest terms. It can be shown that $f$ is continuous precisely at the irrational points. It obviously satisfies your condition with $L=1$ because every interval contains distinct irrational numbers. However, it cannot be uniformly continuous, because it is not continuous.

share|cite|improve this answer

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

$endgroup$

share|cite|improve this answer

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

answered Mar 18 at 13:47

uniquesolutionuniquesolution

9,4471823

1

$begingroup$

The non-uniformly continuous function, $C^1$ function $f(x) = x sin x$ satisfies your criterion, because it has a sequence of critical points $x_n$ that approach infinity. For any critical point $x$ of any $C^1$ function $f$ and for any value of $L$, there exists $delta>0$ such that your inequality is true for all $x_1,x_2 in (x-delta,x+delta)$.

When I think of the $L$-Lipschitz condition, I like to remember one of its properties: if $f$ is differentiable and $L$-Lipschitz then $|f'(x)| le L$ for all $x$ in the domain of $f$. I concocted my example so that $f(x)$ has two phenomena: many critical points, and points with larger and larger derivative.

share|cite|improve this answer

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

$endgroup$

add a comment | 

1

$begingroup$

The non-uniformly continuous function, $C^1$ function $f(x) = x sin x$ satisfies your criterion, because it has a sequence of critical points $x_n$ that approach infinity. For any critical point $x$ of any $C^1$ function $f$ and for any value of $L$, there exists $delta>0$ such that your inequality is true for all $x_1,x_2 in (x-delta,x+delta)$.

When I think of the $L$-Lipschitz condition, I like to remember one of its properties: if $f$ is differentiable and $L$-Lipschitz then $|f'(x)| le L$ for all $x$ in the domain of $f$. I concocted my example so that $f(x)$ has two phenomena: many critical points, and points with larger and larger derivative.

share|cite|improve this answer

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

$endgroup$

add a comment | 

$begingroup$

The non-uniformly continuous function, $C^1$ function $f(x) = x sin x$ satisfies your criterion, because it has a sequence of critical points $x_n$ that approach infinity. For any critical point $x$ of any $C^1$ function $f$ and for any value of $L$, there exists $delta>0$ such that your inequality is true for all $x_1,x_2 in (x-delta,x+delta)$.

When I think of the $L$-Lipschitz condition, I like to remember one of its properties: if $f$ is differentiable and $L$-Lipschitz then $|f'(x)| le L$ for all $x$ in the domain of $f$. I concocted my example so that $f(x)$ has two phenomena: many critical points, and points with larger and larger derivative.

share|cite|improve this answer

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

$endgroup$

share|cite|improve this answer

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889

answered Mar 18 at 13:47

Lee MosherLee Mosher

51.4k33889