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For which $ainmathbbR$ is $f: (0,infty)tomathbbR$, $f(x)=x^a$ Lipschitz-continuous?
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I want to give all $ainmathbbR$, for which $f_a: (0,infty)tomathbbR$, $f_a(x)=x^a$ is Lipschitz-continuous?
I think, that $f_a$ is only Lipschitz-continuous for $a=0$ and $a=1$, which is trivial.
But what is the best way, to show, that it is not Lipschitz-continuous for every other $a$, using the definition only?
I want to seperate three cases. $ain (-infty, 0)$, $ain (0,1)$ and $ain (1,infty)$
For the first case $ain (-infty, 0)$ I just observe $(0,1]$. Because, when I can show, that it fails on this interval, $f_a$ can not be Lipschitz.
I have to show, that $forall Lgeq 0quadexists x,yin (0,1]: |f_a(x)-f_a(y)|> L|x-y|$
Without loss of generality, we can take $y=1$ and get $|x^a-1|=x^a-1$, since $xin (0,1]$.
Also if we can show, that $x^a-1>L$, then $x^a-1> L|x-y|$
This gives us $x^a>L+1Leftrightarrow x^-a<frac1L+1$.
Taking the "(-a)th root", leaves us with $x<sqrt[-a]frac1L+1$.
Now I choose, lets say $x=sqrt[-a]frac1L+2$ and $y=1$ and conclude the proof.
Would this be sound? Taking the "(-a)th root" feels not that good. Also, can the choice of $x$ depend on $a$ too? I think so, since $a$ is fixed.
Is this correct so far?
Thanks in advance.
proof-verification continuity proof-writing lipschitz-functions
asked Aug 1 at 2:40
Cornman
2,29021027
add a comment |Â
up vote
1
down vote
favorite
I want to give all $ainmathbbR$, for which $f_a: (0,infty)tomathbbR$, $f_a(x)=x^a$ is Lipschitz-continuous?
I think, that $f_a$ is only Lipschitz-continuous for $a=0$ and $a=1$, which is trivial.
But what is the best way, to show, that it is not Lipschitz-continuous for every other $a$, using the definition only?
I want to seperate three cases. $ain (-infty, 0)$, $ain (0,1)$ and $ain (1,infty)$
For the first case $ain (-infty, 0)$ I just observe $(0,1]$. Because, when I can show, that it fails on this interval, $f_a$ can not be Lipschitz.
I have to show, that $forall Lgeq 0quadexists x,yin (0,1]: |f_a(x)-f_a(y)|> L|x-y|$
Without loss of generality, we can take $y=1$ and get $|x^a-1|=x^a-1$, since $xin (0,1]$.
Also if we can show, that $x^a-1>L$, then $x^a-1> L|x-y|$
This gives us $x^a>L+1Leftrightarrow x^-a<frac1L+1$.
Taking the "(-a)th root", leaves us with $x<sqrt[-a]frac1L+1$.
Now I choose, lets say $x=sqrt[-a]frac1L+2$ and $y=1$ and conclude the proof.
Would this be sound? Taking the "(-a)th root" feels not that good. Also, can the choice of $x$ depend on $a$ too? I think so, since $a$ is fixed.
Is this correct so far?
Thanks in advance.
proof-verification continuity proof-writing lipschitz-functions
asked Aug 1 at 2:40
Cornman
2,29021027
2
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to give all $ainmathbbR$, for which $f_a: (0,infty)tomathbbR$, $f_a(x)=x^a$ is Lipschitz-continuous?
I think, that $f_a$ is only Lipschitz-continuous for $a=0$ and $a=1$, which is trivial.
But what is the best way, to show, that it is not Lipschitz-continuous for every other $a$, using the definition only?
I want to seperate three cases. $ain (-infty, 0)$, $ain (0,1)$ and $ain (1,infty)$
For the first case $ain (-infty, 0)$ I just observe $(0,1]$. Because, when I can show, that it fails on this interval, $f_a$ can not be Lipschitz.
I have to show, that $forall Lgeq 0quadexists x,yin (0,1]: |f_a(x)-f_a(y)|> L|x-y|$
Without loss of generality, we can take $y=1$ and get $|x^a-1|=x^a-1$, since $xin (0,1]$.
Also if we can show, that $x^a-1>L$, then $x^a-1> L|x-y|$
This gives us $x^a>L+1Leftrightarrow x^-a<frac1L+1$.
Taking the "(-a)th root", leaves us with $x<sqrt[-a]frac1L+1$.
Now I choose, lets say $x=sqrt[-a]frac1L+2$ and $y=1$ and conclude the proof.
Would this be sound? Taking the "(-a)th root" feels not that good. Also, can the choice of $x$ depend on $a$ too? I think so, since $a$ is fixed.
Is this correct so far?
Thanks in advance.
proof-verification continuity proof-writing lipschitz-functions
asked Aug 1 at 2:40
Cornman
2,29021027
I want to give all $ainmathbbR$, for which $f_a: (0,infty)tomathbbR$, $f_a(x)=x^a$ is Lipschitz-continuous?
I think, that $f_a$ is only Lipschitz-continuous for $a=0$ and $a=1$, which is trivial.
But what is the best way, to show, that it is not Lipschitz-continuous for every other $a$, using the definition only?
I want to seperate three cases. $ain (-infty, 0)$, $ain (0,1)$ and $ain (1,infty)$
For the first case $ain (-infty, 0)$ I just observe $(0,1]$. Because, when I can show, that it fails on this interval, $f_a$ can not be Lipschitz.
I have to show, that $forall Lgeq 0quadexists x,yin (0,1]: |f_a(x)-f_a(y)|> L|x-y|$
Without loss of generality, we can take $y=1$ and get $|x^a-1|=x^a-1$, since $xin (0,1]$.
Also if we can show, that $x^a-1>L$, then $x^a-1> L|x-y|$
This gives us $x^a>L+1Leftrightarrow x^-a<frac1L+1$.
Taking the "(-a)th root", leaves us with $x<sqrt[-a]frac1L+1$.
Now I choose, lets say $x=sqrt[-a]frac1L+2$ and $y=1$ and conclude the proof.
Would this be sound? Taking the "(-a)th root" feels not that good. Also, can the choice of $x$ depend on $a$ too? I think so, since $a$ is fixed.
Is this correct so far?
Thanks in advance.
proof-verification continuity proof-writing lipschitz-functions
asked Aug 1 at 2:40
Cornman
2,29021027
edited Aug 1 at 2:51
asked Aug 1 at 2:40
Cornman
2,29021027
asked Aug 1 at 2:40
Cornman
2,29021027
asked Aug 1 at 2:40
Cornman
2,29021027
2,29021027
2
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35
add a comment |Â
2
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35
2
2
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Your conclusion is right: here's another look:
If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $forall x neq y$ in the domain we have $$ Bigg|fracf(x)-f(y)x-y Bigg|leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $xin 0,1$.
$|fracx^n-y^nx-y|leq C$ but then if we consider $x=y+1$ this becomes
$(y+1)^n-y^n =y^n bigg( big(fracy+1y big)^n-1 bigg)$ and this is unbounded when $nnotin 0,1$.
answered Aug 1 at 2:58
Mason
1,1341223
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
add a comment |Â
up vote
1
down vote
$f$ is Lipschitz on $Bbb R^+$ iff $sup_(0<x<y)|g(x,y)|<infty$ where $g(x,y)=frac f(y)-f(x)y-x.$
For $0ne ane 1:$ Consider that when $x>0$ we have $$g(x,2x)=frac f(2x)-f(x)2x-x=frac (2x)^a-x^ax=(2^a-1)x^a-1.$$ Since $ane 0implies 2^a-1ne 0,$ it suffices to show that
(i). If $a>1$ then $sup_(xgeq1)x^a-1=infty.$
(ii). If $0ne a<1$ then $sup_(0<xleq 1)x^a-1=infty.$
To prove (i), if $a>1$ there exist $b,c in Bbb Z^+$ with $a-1>b/c.$ Now for any $nin Bbb Z^+$ we have $(n^c)^a-1geq(n^c)^b/c=n^bgeq n.$ Therefore $$sup_xgeq 1 x^a-1geq sup_nin Bbb Z^+(n^c)^a-1geq sup_nin Bbb Z^+n=infty$$ because every $rin Bbb R$ is less than some $nin Bbb Z^+.$
To prove (ii), if $0ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$sup_0<xleq 1x^a-1 =sup_x'geq 1;(1/x')^a-1=sup_x'geq1;(x')^a'-1=infty.$$
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your conclusion is right: here's another look:
If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $forall x neq y$ in the domain we have $$ Bigg|fracf(x)-f(y)x-y Bigg|leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $xin 0,1$.
$|fracx^n-y^nx-y|leq C$ but then if we consider $x=y+1$ this becomes
$(y+1)^n-y^n =y^n bigg( big(fracy+1y big)^n-1 bigg)$ and this is unbounded when $nnotin 0,1$.
answered Aug 1 at 2:58
Mason
1,1341223
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
add a comment |Â
up vote
1
down vote
accepted
Your conclusion is right: here's another look:
If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $forall x neq y$ in the domain we have $$ Bigg|fracf(x)-f(y)x-y Bigg|leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $xin 0,1$.
$|fracx^n-y^nx-y|leq C$ but then if we consider $x=y+1$ this becomes
$(y+1)^n-y^n =y^n bigg( big(fracy+1y big)^n-1 bigg)$ and this is unbounded when $nnotin 0,1$.
answered Aug 1 at 2:58
Mason
1,1341223
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your conclusion is right: here's another look:
If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $forall x neq y$ in the domain we have $$ Bigg|fracf(x)-f(y)x-y Bigg|leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $xin 0,1$.
$|fracx^n-y^nx-y|leq C$ but then if we consider $x=y+1$ this becomes
$(y+1)^n-y^n =y^n bigg( big(fracy+1y big)^n-1 bigg)$ and this is unbounded when $nnotin 0,1$.
answered Aug 1 at 2:58
Mason
1,1341223
Your conclusion is right: here's another look:
If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $forall x neq y$ in the domain we have $$ Bigg|fracf(x)-f(y)x-y Bigg|leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $xin 0,1$.
$|fracx^n-y^nx-y|leq C$ but then if we consider $x=y+1$ this becomes
$(y+1)^n-y^n =y^n bigg( big(fracy+1y big)^n-1 bigg)$ and this is unbounded when $nnotin 0,1$.
answered Aug 1 at 2:58
Mason
1,1341223
edited Aug 1 at 3:44
answered Aug 1 at 2:58
Mason
1,1341223
answered Aug 1 at 2:58
Mason
1,1341223
answered Aug 1 at 2:58
Mason
1,1341223
1,1341223
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
add a comment |Â
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
1
1
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
I appreciate your answer, but I do not want to involve the derivative. Otherwise, what are equivalent properties for Lipschitz-continuous? Can we say a function is Lipschitz iff the derivative is bounded?
– Cornman
Aug 1 at 3:01
1
1
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
How about now? I was really only using this $f'$ as a shorthand for $(f(x)-f(y)) /(x-y)$
– Mason
Aug 1 at 3:24
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
@Cornman, check out the properties on the wiki. I think you are looking for the top list item under properties. We can't quite say in your formulation because Lipshitz-continuity doesn't imply it's differentiable over the entire domain. As for a critique of your proof: I think everything looks fine. I think I might argue that the 3 cases is needlessly verbose (but that's not really a problem).
– Mason
Aug 1 at 15:08
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
I really like your answer now. Did not notice your "shorthand use" of $f'$. :)
– Cornman
Aug 1 at 15:12
add a comment |Â
up vote
1
down vote
$f$ is Lipschitz on $Bbb R^+$ iff $sup_(0<x<y)|g(x,y)|<infty$ where $g(x,y)=frac f(y)-f(x)y-x.$
For $0ne ane 1:$ Consider that when $x>0$ we have $$g(x,2x)=frac f(2x)-f(x)2x-x=frac (2x)^a-x^ax=(2^a-1)x^a-1.$$ Since $ane 0implies 2^a-1ne 0,$ it suffices to show that
(i). If $a>1$ then $sup_(xgeq1)x^a-1=infty.$
(ii). If $0ne a<1$ then $sup_(0<xleq 1)x^a-1=infty.$
To prove (i), if $a>1$ there exist $b,c in Bbb Z^+$ with $a-1>b/c.$ Now for any $nin Bbb Z^+$ we have $(n^c)^a-1geq(n^c)^b/c=n^bgeq n.$ Therefore $$sup_xgeq 1 x^a-1geq sup_nin Bbb Z^+(n^c)^a-1geq sup_nin Bbb Z^+n=infty$$ because every $rin Bbb R$ is less than some $nin Bbb Z^+.$
To prove (ii), if $0ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$sup_0<xleq 1x^a-1 =sup_x'geq 1;(1/x')^a-1=sup_x'geq1;(x')^a'-1=infty.$$
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
add a comment |Â
up vote
1
down vote
$f$ is Lipschitz on $Bbb R^+$ iff $sup_(0<x<y)|g(x,y)|<infty$ where $g(x,y)=frac f(y)-f(x)y-x.$
For $0ne ane 1:$ Consider that when $x>0$ we have $$g(x,2x)=frac f(2x)-f(x)2x-x=frac (2x)^a-x^ax=(2^a-1)x^a-1.$$ Since $ane 0implies 2^a-1ne 0,$ it suffices to show that
(i). If $a>1$ then $sup_(xgeq1)x^a-1=infty.$
(ii). If $0ne a<1$ then $sup_(0<xleq 1)x^a-1=infty.$
To prove (i), if $a>1$ there exist $b,c in Bbb Z^+$ with $a-1>b/c.$ Now for any $nin Bbb Z^+$ we have $(n^c)^a-1geq(n^c)^b/c=n^bgeq n.$ Therefore $$sup_xgeq 1 x^a-1geq sup_nin Bbb Z^+(n^c)^a-1geq sup_nin Bbb Z^+n=infty$$ because every $rin Bbb R$ is less than some $nin Bbb Z^+.$
To prove (ii), if $0ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$sup_0<xleq 1x^a-1 =sup_x'geq 1;(1/x')^a-1=sup_x'geq1;(x')^a'-1=infty.$$
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$f$ is Lipschitz on $Bbb R^+$ iff $sup_(0<x<y)|g(x,y)|<infty$ where $g(x,y)=frac f(y)-f(x)y-x.$
For $0ne ane 1:$ Consider that when $x>0$ we have $$g(x,2x)=frac f(2x)-f(x)2x-x=frac (2x)^a-x^ax=(2^a-1)x^a-1.$$ Since $ane 0implies 2^a-1ne 0,$ it suffices to show that
(i). If $a>1$ then $sup_(xgeq1)x^a-1=infty.$
(ii). If $0ne a<1$ then $sup_(0<xleq 1)x^a-1=infty.$
To prove (i), if $a>1$ there exist $b,c in Bbb Z^+$ with $a-1>b/c.$ Now for any $nin Bbb Z^+$ we have $(n^c)^a-1geq(n^c)^b/c=n^bgeq n.$ Therefore $$sup_xgeq 1 x^a-1geq sup_nin Bbb Z^+(n^c)^a-1geq sup_nin Bbb Z^+n=infty$$ because every $rin Bbb R$ is less than some $nin Bbb Z^+.$
To prove (ii), if $0ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$sup_0<xleq 1x^a-1 =sup_x'geq 1;(1/x')^a-1=sup_x'geq1;(x')^a'-1=infty.$$
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
$f$ is Lipschitz on $Bbb R^+$ iff $sup_(0<x<y)|g(x,y)|<infty$ where $g(x,y)=frac f(y)-f(x)y-x.$
For $0ne ane 1:$ Consider that when $x>0$ we have $$g(x,2x)=frac f(2x)-f(x)2x-x=frac (2x)^a-x^ax=(2^a-1)x^a-1.$$ Since $ane 0implies 2^a-1ne 0,$ it suffices to show that
(i). If $a>1$ then $sup_(xgeq1)x^a-1=infty.$
(ii). If $0ne a<1$ then $sup_(0<xleq 1)x^a-1=infty.$
To prove (i), if $a>1$ there exist $b,c in Bbb Z^+$ with $a-1>b/c.$ Now for any $nin Bbb Z^+$ we have $(n^c)^a-1geq(n^c)^b/c=n^bgeq n.$ Therefore $$sup_xgeq 1 x^a-1geq sup_nin Bbb Z^+(n^c)^a-1geq sup_nin Bbb Z^+n=infty$$ because every $rin Bbb R$ is less than some $nin Bbb Z^+.$
To prove (ii), if $0ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$sup_0<xleq 1x^a-1 =sup_x'geq 1;(1/x')^a-1=sup_x'geq1;(x')^a'-1=infty.$$
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
edited Aug 1 at 5:42
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
answered Aug 1 at 5:09
DanielWainfleet
31.4k31542
31.4k31542
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
add a comment |Â
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
I know I have not critiqued the proposer's proof. I wanted to present a model of how to prove it.
– DanielWainfleet
Aug 1 at 5:50
add a comment |Â
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2
Just to add some intuition: A function $f(x)$ is not Lipschitz on its domain if its derivative is unbounded there. Thus if $a > 1$, then $f_a'(x) to infty$ as $x to infty$. If $a in (-infty, 0) cup (0,1)$, then $f_a'(x) to pm infty$ as $x to 0^+$.
– JavaMan
Aug 1 at 2:52
@JavaMan Yes, I know that. But I would like to just use the definition.
– Cornman
Aug 1 at 2:54
Are differential calculus, and the MVT, not allowed?
– DanielWainfleet
Aug 1 at 4:39
@DanielWainfleet It is allowed, but I would like not too. But feel free, to post an answer.
– Cornman
Aug 1 at 13:35