Mixing
Let $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$.
$begingroup$
Let $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$. Also, I want to know if $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$.
My attemp of proof goes as follow, let $f in Lip_{beta}[a,b]$, then for every $x,y in [a,b]$ I got that there is a $M>0$ such $|f(x)-f(y)| leq M|x-y|^{beta}$. As someone point me below in the comments, I have that
$$|f(x)-f(y)| leq M|x-y|^{beta}=M|x-y|^{beta-alpha}|x-y|^{alpha}.$$
So I think the $M' > 0$ im looking for is $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$, this way for every $x,y in [a,b]$ there is an $M>0$ such that
$$|f(x)-f(y)|leq M|x-y|^{beta}=M|x-y|^{beta- alpha }|x-y|^{alpha} leq sup lbrace M|x-y|^{beta- alpha} rbrace=M'|x-y|^{alpha}.$$
Is my proof right?
For $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$ I was thinking in using the equivalence of a closed subset as a subset which contains all its limit points. Then how do I proof this subset contains all its limit point, Im working here with the supremum norm of the space of continuous functions. Thank you!
real-analysis calculus general-topology functional-analysis analysis
edited Feb 6 at 2:54
d.k.o.
10.5k630
asked Jan 29 at 1:53
CosCos
24028
$endgroup$
|
show 1 more comment
$begingroup$
Let $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$. Also, I want to know if $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$.
My attemp of proof goes as follow, let $f in Lip_{beta}[a,b]$, then for every $x,y in [a,b]$ I got that there is a $M>0$ such $|f(x)-f(y)| leq M|x-y|^{beta}$. As someone point me below in the comments, I have that
$$|f(x)-f(y)| leq M|x-y|^{beta}=M|x-y|^{beta-alpha}|x-y|^{alpha}.$$
So I think the $M' > 0$ im looking for is $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$, this way for every $x,y in [a,b]$ there is an $M>0$ such that
$$|f(x)-f(y)|leq M|x-y|^{beta}=M|x-y|^{beta- alpha }|x-y|^{alpha} leq sup lbrace M|x-y|^{beta- alpha} rbrace=M'|x-y|^{alpha}.$$
Is my proof right?
For $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$ I was thinking in using the equivalence of a closed subset as a subset which contains all its limit points. Then how do I proof this subset contains all its limit point, Im working here with the supremum norm of the space of continuous functions. Thank you!
real-analysis calculus general-topology functional-analysis analysis
edited Feb 6 at 2:54
d.k.o.
10.5k630
asked Jan 29 at 1:53
CosCos
24028
$endgroup$
$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58
|
show 1 more comment
$begingroup$
Let $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$. Also, I want to know if $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$.
My attemp of proof goes as follow, let $f in Lip_{beta}[a,b]$, then for every $x,y in [a,b]$ I got that there is a $M>0$ such $|f(x)-f(y)| leq M|x-y|^{beta}$. As someone point me below in the comments, I have that
$$|f(x)-f(y)| leq M|x-y|^{beta}=M|x-y|^{beta-alpha}|x-y|^{alpha}.$$
So I think the $M' > 0$ im looking for is $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$, this way for every $x,y in [a,b]$ there is an $M>0$ such that
$$|f(x)-f(y)|leq M|x-y|^{beta}=M|x-y|^{beta- alpha }|x-y|^{alpha} leq sup lbrace M|x-y|^{beta- alpha} rbrace=M'|x-y|^{alpha}.$$
Is my proof right?
For $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$ I was thinking in using the equivalence of a closed subset as a subset which contains all its limit points. Then how do I proof this subset contains all its limit point, Im working here with the supremum norm of the space of continuous functions. Thank you!
real-analysis calculus general-topology functional-analysis analysis
edited Feb 6 at 2:54
d.k.o.
10.5k630
asked Jan 29 at 1:53
CosCos
24028
$endgroup$
Let $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$. Also, I want to know if $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$.
My attemp of proof goes as follow, let $f in Lip_{beta}[a,b]$, then for every $x,y in [a,b]$ I got that there is a $M>0$ such $|f(x)-f(y)| leq M|x-y|^{beta}$. As someone point me below in the comments, I have that
$$|f(x)-f(y)| leq M|x-y|^{beta}=M|x-y|^{beta-alpha}|x-y|^{alpha}.$$
So I think the $M' > 0$ im looking for is $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$, this way for every $x,y in [a,b]$ there is an $M>0$ such that
$$|f(x)-f(y)|leq M|x-y|^{beta}=M|x-y|^{beta- alpha }|x-y|^{alpha} leq sup lbrace M|x-y|^{beta- alpha} rbrace=M'|x-y|^{alpha}.$$
Is my proof right?
For $Lip_beta[a,b]$ is a closed subset for $Lip_{alpha}[a,b]$ I was thinking in using the equivalence of a closed subset as a subset which contains all its limit points. Then how do I proof this subset contains all its limit point, Im working here with the supremum norm of the space of continuous functions. Thank you!
real-analysis calculus general-topology functional-analysis analysis
real-analysis calculus general-topology functional-analysis analysis
edited Feb 6 at 2:54
d.k.o.
10.5k630
asked Jan 29 at 1:53
CosCos
24028
edited Feb 6 at 2:54
d.k.o.
10.5k630
asked Jan 29 at 1:53
CosCos
24028
edited Feb 6 at 2:54
d.k.o.
10.5k630
edited Feb 6 at 2:54
d.k.o.
10.5k630
edited Feb 6 at 2:54
d.k.o.
10.5k630
10.5k630
asked Jan 29 at 1:53
CosCos
24028
asked Jan 29 at 1:53
CosCos
24028
asked Jan 29 at 1:53
CosCos
24028
24028
$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58
|
show 1 more comment
$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58
$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$operatorname{Lip}_{beta}[a,b]$ is not closed in $operatorname{Lip}_{alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $alpha=1/2$, and $beta=1$. Consider $f(x)=sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}vee f$. Then each $f_n$ is Lipschitz and $|f_n-f|_{infty}=n^{-1}to 0$ as $ntoinfty$. However, $fnotin operatorname{Lip}_1[0,1]$.
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
add a comment |
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1 Answer
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$begingroup$
$operatorname{Lip}_{beta}[a,b]$ is not closed in $operatorname{Lip}_{alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $alpha=1/2$, and $beta=1$. Consider $f(x)=sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}vee f$. Then each $f_n$ is Lipschitz and $|f_n-f|_{infty}=n^{-1}to 0$ as $ntoinfty$. However, $fnotin operatorname{Lip}_1[0,1]$.
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
add a comment |
$begingroup$
$operatorname{Lip}_{beta}[a,b]$ is not closed in $operatorname{Lip}_{alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $alpha=1/2$, and $beta=1$. Consider $f(x)=sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}vee f$. Then each $f_n$ is Lipschitz and $|f_n-f|_{infty}=n^{-1}to 0$ as $ntoinfty$. However, $fnotin operatorname{Lip}_1[0,1]$.
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
add a comment |
$begingroup$
$operatorname{Lip}_{beta}[a,b]$ is not closed in $operatorname{Lip}_{alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $alpha=1/2$, and $beta=1$. Consider $f(x)=sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}vee f$. Then each $f_n$ is Lipschitz and $|f_n-f|_{infty}=n^{-1}to 0$ as $ntoinfty$. However, $fnotin operatorname{Lip}_1[0,1]$.
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
$endgroup$
$operatorname{Lip}_{beta}[a,b]$ is not closed in $operatorname{Lip}_{alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $alpha=1/2$, and $beta=1$. Consider $f(x)=sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}vee f$. Then each $f_n$ is Lipschitz and $|f_n-f|_{infty}=n^{-1}to 0$ as $ntoinfty$. However, $fnotin operatorname{Lip}_1[0,1]$.
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
edited Feb 6 at 6:00
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
answered Feb 6 at 2:06
d.k.o.d.k.o.
10.5k630
10.5k630
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
add a comment |
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/
$endgroup$
– Cos
Mar 2 at 21:52
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
$begingroup$
@Cos $avee bequiv max{a,b}$.
$endgroup$
– d.k.o.
Mar 2 at 22:30
add a comment |
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$begingroup$
$$ M|x-y|^{beta}= M|x-y|^{alpha}|x-y|^{beta-alpha}le M'|x-y|^{alpha}. $$
$endgroup$
– d.k.o.
Jan 29 at 2:59
$begingroup$
What is your definition of $Lip_alpha[a,b]$?
$endgroup$
– d.k.o.
Jan 29 at 3:00
$begingroup$
$f in Lip_{alpha}[a,b]$ if for every $x,y in [a,b]$, $|f(x)-f(y)| leq M|x-y|^{alpha}$ for some $M>0$. @d.k.o.
$endgroup$
– Cos
Jan 29 at 3:02
$begingroup$
@d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup lbrace M|x-y|^{beta-alpha} rbrace$ isnt?
$endgroup$
– Cos
Feb 6 at 1:40
$begingroup$
Yes, and this supremum can be computed.
$endgroup$
– d.k.o.
Feb 6 at 1:58