Given an averaging operator $A: l_p rightarrow l_p$, why $A$ is not compact












2












$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56
















2












$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56














2












2








2


1



$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$




Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators




functional-analysis operator-theory compactness lp-spaces compact-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:57









Davide Giraudo

127k16151263




127k16151263










asked Dec 9 '18 at 17:10









Anton ZagrivinAnton Zagrivin

1548




1548












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56


















  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56
















$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 9 '18 at 17:11




$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 9 '18 at 17:11












$begingroup$
I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:30




$begingroup$
I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:30












$begingroup$
I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32




$begingroup$
I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32












$begingroup$
Maybe my entire idea is wrong and it will not work here
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32




$begingroup$
Maybe my entire idea is wrong and it will not work here
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32












$begingroup$
A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
$endgroup$
– Jean Marie
Dec 9 '18 at 17:56




$begingroup$
A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
$endgroup$
– Jean Marie
Dec 9 '18 at 17:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22
















1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22














1












1








1





$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$



Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 9:57

























answered Dec 9 '18 at 22:28









Davide GiraudoDavide Giraudo

127k16151263




127k16151263












  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22


















  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22
















$begingroup$
Oh, that's what I was trying to do but couldn't achieve
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16




$begingroup$
Oh, that's what I was trying to do but couldn't achieve
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16












$begingroup$
Thank you a lot
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16




$begingroup$
Thank you a lot
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16












$begingroup$
You are welcome
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:22




$begingroup$
You are welcome
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:22


















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