$A_1,A_2$ fulfill property, but their sum $A_1+A_2$ does not












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 12:11







Tesla

















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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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