Mixing
Equivalence between Hölder norms
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Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?
calculus real-analysis functional-analysis holder-spaces holder-inequality
edited Jul 28 at 17:37
copper.hat
122k557156
asked Jul 28 at 17:07
JCF
18411
add a comment |Â
up vote
0
down vote
favorite
Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?
calculus real-analysis functional-analysis holder-spaces holder-inequality
edited Jul 28 at 17:37
copper.hat
122k557156
asked Jul 28 at 17:07
JCF
18411
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?
calculus real-analysis functional-analysis holder-spaces holder-inequality
edited Jul 28 at 17:37
copper.hat
122k557156
asked Jul 28 at 17:07
JCF
18411
Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?
calculus real-analysis functional-analysis holder-spaces holder-inequality
edited Jul 28 at 17:37
copper.hat
122k557156
asked Jul 28 at 17:07
JCF
18411
edited Jul 28 at 17:37
copper.hat
122k557156
edited Jul 28 at 17:37
copper.hat
122k557156
edited Jul 28 at 17:37
copper.hat
122k557156
122k557156
asked Jul 28 at 17:07
JCF
18411
asked Jul 28 at 17:07
JCF
18411
asked Jul 28 at 17:07
JCF
18411
18411
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.
The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.
answered Jul 28 at 17:30
copper.hat
122k557156
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
add a comment |Â
up vote
2
down vote
It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.
For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.
answered Jul 28 at 17:30
Rhys Steele
5,5651828
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.
The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.
answered Jul 28 at 17:30
copper.hat
122k557156
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
add a comment |Â
up vote
2
down vote
accepted
begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.
The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.
answered Jul 28 at 17:30
copper.hat
122k557156
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.
The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.
answered Jul 28 at 17:30
copper.hat
122k557156
begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.
The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.
answered Jul 28 at 17:30
copper.hat
122k557156
answered Jul 28 at 17:30
copper.hat
122k557156
answered Jul 28 at 17:30
copper.hat
122k557156
answered Jul 28 at 17:30
copper.hat
122k557156
122k557156
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
add a comment |Â
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20
add a comment |Â
up vote
2
down vote
It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.
For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.
answered Jul 28 at 17:30
Rhys Steele
5,5651828
add a comment |Â
up vote
2
down vote
It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.
For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.
answered Jul 28 at 17:30
Rhys Steele
5,5651828
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.
For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.
answered Jul 28 at 17:30
Rhys Steele
5,5651828
It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.
For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.
answered Jul 28 at 17:30
Rhys Steele
5,5651828
edited Jul 28 at 18:42
answered Jul 28 at 17:30
Rhys Steele
5,5651828
answered Jul 28 at 17:30
Rhys Steele
5,5651828
answered Jul 28 at 17:30
Rhys Steele
5,5651828
5,5651828
add a comment |Â
add a comment |Â
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