$A_1,A_2$ fulfill property, but their sum $A_1+A_2$ does not
$begingroup$
Let $V$ be a real, reflexive, separable Banach space.
Are there operators $A_1,A_2: V to V^*$ that fulfill the property
begin{cases}
u_n rightharpoonup u \
A_iu_n rightharpoonup b \
limsup_{n to infty} langle A_iu_n,u_nrangleleqlangle b,urangle
end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n in mathbb N}$ in $V$, $b in V^*$, but their sum $A_1+A_2$ doesn't?
My work:
$(A_1+A_2)u_n=A_1u_n+A_2u_n rightharpoonup2b in V^*$ and
$limsup, langle (A_1+A_2)u_n,u_n rangle=limsup langle A_1u_n,u_n rangle+limsup langle A_2u_n,u_n rangle leq 2 langle b,urangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u neq 2b?$
functional-analysis operator-theory banach-spaces weak-convergence
asked Dec 12 '18 at 9:26
TeslaTesla
890426
$endgroup$
|
show 2 more comments
$begingroup$
Let $V$ be a real, reflexive, separable Banach space.
Are there operators $A_1,A_2: V to V^*$ that fulfill the property
begin{cases}
u_n rightharpoonup u \
A_iu_n rightharpoonup b \
limsup_{n to infty} langle A_iu_n,u_nrangleleqlangle b,urangle
end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n in mathbb N}$ in $V$, $b in V^*$, but their sum $A_1+A_2$ doesn't?
My work:
$(A_1+A_2)u_n=A_1u_n+A_2u_n rightharpoonup2b in V^*$ and
$limsup, langle (A_1+A_2)u_n,u_n rangle=limsup langle A_1u_n,u_n rangle+limsup langle A_2u_n,u_n rangle leq 2 langle b,urangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u neq 2b?$
functional-analysis operator-theory banach-spaces weak-convergence
asked Dec 12 '18 at 9:26
TeslaTesla
890426
$endgroup$
1
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40
|
show 2 more comments
$begingroup$
Let $V$ be a real, reflexive, separable Banach space.
Are there operators $A_1,A_2: V to V^*$ that fulfill the property
begin{cases}
u_n rightharpoonup u \
A_iu_n rightharpoonup b \
limsup_{n to infty} langle A_iu_n,u_nrangleleqlangle b,urangle
end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n in mathbb N}$ in $V$, $b in V^*$, but their sum $A_1+A_2$ doesn't?
My work:
$(A_1+A_2)u_n=A_1u_n+A_2u_n rightharpoonup2b in V^*$ and
$limsup, langle (A_1+A_2)u_n,u_n rangle=limsup langle A_1u_n,u_n rangle+limsup langle A_2u_n,u_n rangle leq 2 langle b,urangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u neq 2b?$
functional-analysis operator-theory banach-spaces weak-convergence
asked Dec 12 '18 at 9:26
TeslaTesla
890426
$endgroup$
Let $V$ be a real, reflexive, separable Banach space.
Are there operators $A_1,A_2: V to V^*$ that fulfill the property
begin{cases}
u_n rightharpoonup u \
A_iu_n rightharpoonup b \
limsup_{n to infty} langle A_iu_n,u_nrangleleqlangle b,urangle
end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n in mathbb N}$ in $V$, $b in V^*$, but their sum $A_1+A_2$ doesn't?
My work:
$(A_1+A_2)u_n=A_1u_n+A_2u_n rightharpoonup2b in V^*$ and
$limsup, langle (A_1+A_2)u_n,u_n rangle=limsup langle A_1u_n,u_n rangle+limsup langle A_2u_n,u_n rangle leq 2 langle b,urangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u neq 2b?$
functional-analysis operator-theory banach-spaces weak-convergence
functional-analysis operator-theory banach-spaces weak-convergence
asked Dec 12 '18 at 9:26
TeslaTesla
890426
asked Dec 12 '18 at 9:26
TeslaTesla
890426
edited Dec 12 '18 at 12:11
Tesla
asked Dec 12 '18 at 9:26
TeslaTesla
890426
asked Dec 12 '18 at 9:26
TeslaTesla
890426
asked Dec 12 '18 at 9:26
TeslaTesla
890426
890426
1
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40
|
show 2 more comments
1
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40
1
1
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Here is a counterexample. Let $V = mathbb{R}$ and
$$
A_1(x)=
begin{cases}
frac1x & text{if } x ne 0 \
42 & text{if } x = 0
end{cases}
$$
and
$$
A_2(x)=
begin{cases}
-frac1x & text{if } x ne 0 \
23 & text{if } x = 0
end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
$endgroup$
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
add a comment |
$begingroup$
No, this is not possible.
Using the linearity of $A$ in these properties and an inequality involving $limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
$endgroup$
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a counterexample. Let $V = mathbb{R}$ and
$$
A_1(x)=
begin{cases}
frac1x & text{if } x ne 0 \
42 & text{if } x = 0
end{cases}
$$
and
$$
A_2(x)=
begin{cases}
-frac1x & text{if } x ne 0 \
23 & text{if } x = 0
end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
$endgroup$
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
add a comment |
$begingroup$
Here is a counterexample. Let $V = mathbb{R}$ and
$$
A_1(x)=
begin{cases}
frac1x & text{if } x ne 0 \
42 & text{if } x = 0
end{cases}
$$
and
$$
A_2(x)=
begin{cases}
-frac1x & text{if } x ne 0 \
23 & text{if } x = 0
end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
$endgroup$
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
add a comment |
$begingroup$
Here is a counterexample. Let $V = mathbb{R}$ and
$$
A_1(x)=
begin{cases}
frac1x & text{if } x ne 0 \
42 & text{if } x = 0
end{cases}
$$
and
$$
A_2(x)=
begin{cases}
-frac1x & text{if } x ne 0 \
23 & text{if } x = 0
end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
$endgroup$
Here is a counterexample. Let $V = mathbb{R}$ and
$$
A_1(x)=
begin{cases}
frac1x & text{if } x ne 0 \
42 & text{if } x = 0
end{cases}
$$
and
$$
A_2(x)=
begin{cases}
-frac1x & text{if } x ne 0 \
23 & text{if } x = 0
end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
answered Dec 13 '18 at 8:10
gerwgerw
19.8k11334
19.8k11334
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
add a comment |
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
Hm thank you, but how does u have to be chosen? it does not work for me
$endgroup$
– Tesla
Dec 15 '18 at 21:37
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
$begingroup$
You have to choose $u = 0$.
$endgroup$
– gerw
Dec 16 '18 at 18:56
add a comment |
$begingroup$
No, this is not possible.
Using the linearity of $A$ in these properties and an inequality involving $limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
$endgroup$
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
add a comment |
$begingroup$
No, this is not possible.
Using the linearity of $A$ in these properties and an inequality involving $limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
$endgroup$
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
add a comment |
$begingroup$
No, this is not possible.
Using the linearity of $A$ in these properties and an inequality involving $limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
$endgroup$
No, this is not possible.
Using the linearity of $A$ in these properties and an inequality involving $limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
answered Dec 12 '18 at 9:39
supinfsupinf
6,6921028
6,6921028
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
add a comment |
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
thanks. sorry, i mistyped my question. The operators need not to be bounded/coercive.
$endgroup$
– Tesla
Dec 12 '18 at 10:01
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
I think my answer should still be true
$endgroup$
– supinf
Dec 12 '18 at 10:15
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
$begingroup$
You cannot conclude $A_1u_n$ and $A_2u_n$ to converge weakly if $(A_1+A_2)u_n$ converges weakly.
$endgroup$
– daw
Dec 12 '18 at 19:21
add a comment |
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1
$begingroup$
Any thoughts on the given problem ?
$endgroup$
– Rebellos
Dec 12 '18 at 9:36
$begingroup$
Should this property hold for one sequence $(u_n)$ or for all sequences $(u_n)$?
$endgroup$
– gerw
Dec 12 '18 at 10:38
$begingroup$
@gerw for one sequence $(u_n)$.. and I forgot that the property implies $A_iu=b$..
$endgroup$
– Tesla
Dec 12 '18 at 11:05
$begingroup$
Are the operators $A_i$ assumed to be linear? The claim is trivially true for linear and continuous operators.
$endgroup$
– daw
Dec 12 '18 at 13:42
$begingroup$
No it is not assumed
$endgroup$
– Tesla
Dec 12 '18 at 14:40