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Comparing two summable conditions
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Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:
(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;
(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;
Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).
That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!
functional-analysis analysis banach-spaces examples-counterexamples
asked Nov 14 at 11:34
Greywhite
655
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up vote
3
down vote
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Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:
(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;
(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;
Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).
That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!
functional-analysis analysis banach-spaces examples-counterexamples
asked Nov 14 at 11:34
Greywhite
655
this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:
(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;
(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;
Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).
That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!
functional-analysis analysis banach-spaces examples-counterexamples
asked Nov 14 at 11:34
Greywhite
655
Let $X$ be a Banach space and $(f_n)_{nge1}$ be a sequence in $X^*$. There are two summable conditions:
(1)$sum_{nge1} f_n(x)$ is summable for each x in $X$;
(2)$sum_{nge1} phi(f_n)$ is summable for each $phi$ in $X^{**}$;
Show that (2) implies (1) and (1) doesn't imply (2)(give a counterexample which satisfies (1) but fails (2)).
That (2) implies (1) is trivial. But I don't know how to start with constructing a counterexample. Any help would be appreciated. Thanks in advance!
functional-analysis analysis banach-spaces examples-counterexamples
functional-analysis analysis banach-spaces examples-counterexamples
asked Nov 14 at 11:34
Greywhite
655
asked Nov 14 at 11:34
Greywhite
655
edited 2 days ago
asked Nov 14 at 11:34
Greywhite
655
asked Nov 14 at 11:34
Greywhite
655
asked Nov 14 at 11:34
Greywhite
655
655
this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago
add a comment |
this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago
this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago
add a comment |
2 Answers
2
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up vote
1
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I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:
(1) $sum_{kgeq1}|a_k^n|<infty$;
(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$
(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$
(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$
Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$
one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$
which is finite since $sum_{ngeq1}a_k^n<infty$.
Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$
I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$
whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.
answered 2 days ago
Marco
1908
add a comment |
up vote
1
down vote
I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
$$$$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$
by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.
Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$
You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$
and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.
answered yesterday
Greywhite
655
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:
(1) $sum_{kgeq1}|a_k^n|<infty$;
(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$
(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$
(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$
Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$
one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$
which is finite since $sum_{ngeq1}a_k^n<infty$.
Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$
I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$
whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.
answered 2 days ago
Marco
1908
add a comment |
up vote
1
down vote
I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:
(1) $sum_{kgeq1}|a_k^n|<infty$;
(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$
(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$
(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$
Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$
one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$
which is finite since $sum_{ngeq1}a_k^n<infty$.
Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$
I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$
whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.
answered 2 days ago
Marco
1908
add a comment |
up vote
1
down vote
up vote
1
down vote
I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:
(1) $sum_{kgeq1}|a_k^n|<infty$;
(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$
(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$
(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$
Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$
one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$
which is finite since $sum_{ngeq1}a_k^n<infty$.
Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$
I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$
whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.
answered 2 days ago
Marco
1908
I thought of an example, hopefully is correct.
Of course you need $X$ to be not reflexive. Take $X=c_0$, hence $X^*=ell^1$ and $X^{**}=ell^infty$ (with the usual norms). The functions $f_n$ are represented in $ell^1$ as summable sequences, say $a^n=(a_k^n)_{k=1}^infty$. We have to find $a_k^n$ such that:
(1) $sum_{kgeq1}|a_k^n|<infty$;
(2) For every $(b_k)_{k=1}^inftyin c_0$ it holds
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nb_k<infty
$$
(Remark that the dual pairing between $ell^1$ and $c_0$ is exactly $leftlangle a,brightrangle=sum_{kgeq1}a_kb_k$);
(3) There exists $(c_k)_{k=1}^inftyinell^infty$ such that
$$
sum_{ngeq 1}sum_{kgeq1}a_k^nc_k=infty
$$
(Again, the dual pairing between $ell^infty$ and $ell^1$ is the same as above).
Now, write
$$
a_k^n=frac{1}{n^2+k^2}
$$
Point (1) is trivially true.
For point (2), fix $(b_k)in c_0$ and let $bar{k}$ be such that $|b_k|leq1/n^2$ for $kgeqbar{k}$. Now, since $a_k^ngeq0$ and using the fact that
$$
sum_{kgeqbar{k}}frac{1}{k^2+n^2}|b_k|leqsum_{kgeqbar{k}}frac{1}{k^2}frac{1}{n^2}=Oleft(frac{1}{n^2}right)
$$
one has
$$
left|sum_{ngeq 1}sum_{kgeq1}a_k^nb_kright|leqsum_{ngeq 1}sum_{kgeq1}a_k^n|b_k|
leqsum_{ngeq1}left(sum_{k<bar{k}}a_k^n|b_k|+Oleft(frac{1}{n^2}right)right)
$$
The second term is surely convergent. For the first, using Fubini's theorem, it equals
$$
sum_{k<bar{k}}sum_{ngeq1}a_k^n|b_k|=sum_{k<bar{k}}|b_k|left(sum_{ngeq1}a_k^nright)
$$
which is finite since $sum_{ngeq1}a_k^n<infty$.
Now, for point (3) we could simply choose the element $(1,1,dots)inell^infty$, so that all we need to prove is that
$$
sum_{ngeq 1}sum_{kgeq 1}frac{1}{n^2+k^2}=infty.
$$
I suspected it was true from a comparison with the $mathbb{R}^2$ case, but I asked here in MSE to be sure: Is the double series $sum_{n,kgeq 1}frac{1}{n^2+k^2}$ divergent?.
I hope it works!
P.S. this is what came to me thinking about the problem and I'm quite sure that a simpler (or at least more elegant) example can be built. Choosing to work with $ell^1$, the idea behind this was to find a function $a_k^n=a(n,k)$ defined on $mathbb{N}timesmathbb{N}$, in our case
$$
a(n,k)=frac{1}{n^2+k^2}
$$
whose sections $a(cdot,k_0):mathbb{N}tomathbb{R}$ and $a(n_0,cdot):mathbb{N}tomathbb{R}$ are both integrable (in counting measure, therefore a series) but with $a$ not integrable on the entire $mathbb{N}timesmathbb{N}$.
answered 2 days ago
Marco
1908
edited yesterday
answered 2 days ago
Marco
1908
answered 2 days ago
Marco
1908
answered 2 days ago
Marco
1908
1908
add a comment |
add a comment |
up vote
1
down vote
I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
$$$$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$
by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.
Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$
You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$
and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.
answered yesterday
Greywhite
655
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I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
$$$$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$
by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.
Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$
You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$
and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.
answered yesterday
Greywhite
655
add a comment |
up vote
1
down vote
up vote
1
down vote
I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
$$$$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$
by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.
Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$
You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$
and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.
answered yesterday
Greywhite
655
I came across this problem in a functional analysis textbook by my teacher. The original statement was 'Showing that (1) and (2) are equivalent' and I posted it as it was on MSE. Yesterday I was told this proposition was wrong so I reedited it. Really sorry if my previous post had taken up your time.
I worked out another example. Take $X=ell^1$, hence $X^*=ell^infty$ (with the usual norms). Let $f_ninell^{infty}$ be functionals defined as $$f_{2k-1}^i=0qquad ilt k
$$$$f_{2k-1}^i=1qquad ige k$$$$f_{2k}^i=0qquad ile k$$$$f_{2k}^i=-1qquad igt k$$
That is $$f_1=(1,1,1,1,......)$$$$f_2=(0,-1,-1,-1,......)$$$$f_3=(0,1,1,1,......)$$$$f_4=(0,0,-1,-1,......)$$$$f_5=(0,0,1,1,......)$$$$f_6=(0,0,0,-1,......)$$$$f_7=(0,0,0,1,......)$$$$......$$
Let $(A_n)_{n=1}^{infty}$ be another sequence in $ell^infty$ defined as$$A_n=sum_{k=1}^{n}f_k$$
Then for every $x=(x_i)_{i=1}^inftyin ell^1$ and every $k in mathbb{N}^*$ we have$$A_{2k-1}(x)=sum_{i=1}^{infty}x_i$$$$A_{2k}(x)=sum_{i=1}^{k}x_i$$$$sum_{n=1}^{infty}f_n(x)=lim_{ntoinfty}A_n(x)=sum_{i=1}^{infty}x_i$$
by this we have checked that $sum_{n=1}^{infty}f_n(x)$ is summable for each $x in ell^1$.
Now pick a $phiinell^{infty*}$ so that $sum_{n=1}^{infty}phi(f_n)$ is asummable(Actually this was the most difficult part for me since I don't what $ell^{infty*}$ is exactly). To do this first define $phi$ for $g=(g^i)_{i=1}^{infty}in c$(taken as a subspace of $ell^{infty}$) as$$phi(g)=lim_{itoinfty}g^i$$
You can easily check that $phiin c^*$. By Hahn-Banach Theorem $phi$ has a norm-preserving extension on $l^{infty}$, which we still denote as $phi$. Note that the $f_n$ we defined above are all in $c$. Thus$$phi(f_n)=lim_{itoinfty}f^i=(-1)^n$$
and $sum_{n=1}^{infty}(-1)^n$ is obviously asummable. This finishes the example.
answered yesterday
Greywhite
655
answered yesterday
Greywhite
655
answered yesterday
Greywhite
655
answered yesterday
Greywhite
655
655
add a comment |
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this is just an idea: try a proof by contraposition, that is, suppose that there is some $phiin X^{**}$ such that $sumphi(f_n)$ is not summable, then show that $sum f_n$ cannot be summable also
– Masacroso
Nov 14 at 12:24
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
2 days ago