How to show non existence of an operator $A: V to V^*$

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Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$
functional-analysis operator-theory banach-spaces
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up vote
1
down vote
favorite
Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$
functional-analysis operator-theory banach-spaces
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
1
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$
functional-analysis operator-theory banach-spaces
Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$
functional-analysis operator-theory banach-spaces
functional-analysis operator-theory banach-spaces
edited 6 hours ago
user587192
97810
97810
asked 7 hours ago
Tesla
900426
900426
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
1
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago
|
show 1 more comment
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
1
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
1
1
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Pick an arbitrary point $vin V$ such that $|v|=1$.
Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).
We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.
Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$
For sufficiently large $n$, this cannot be true.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Pick an arbitrary point $vin V$ such that $|v|=1$.
Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).
We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.
Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$
For sufficiently large $n$, this cannot be true.
add a comment |
up vote
1
down vote
accepted
Pick an arbitrary point $vin V$ such that $|v|=1$.
Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).
We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.
Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$
For sufficiently large $n$, this cannot be true.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Pick an arbitrary point $vin V$ such that $|v|=1$.
Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).
We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.
Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$
For sufficiently large $n$, this cannot be true.
Pick an arbitrary point $vin V$ such that $|v|=1$.
Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).
We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.
Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$
For sufficiently large $n$, this cannot be true.
answered 7 hours ago


supinf
5,490926
5,490926
add a comment |
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HSPvBGr,kQERnKwxcl,9,aUs
any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago
No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago
$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago
@LeB its not an inner product, its a duality product
– supinf
7 hours ago
1
@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago