Comparing two summable conditions
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
add a comment |
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
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
Nov 18 at 15:41
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
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
edited Nov 18 at 14:42
asked Nov 14 at 11:34
Greywhite
656
656
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
Nov 18 at 15:41
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
Nov 18 at 15:41
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
Nov 18 at 15:41
Perhaps you could pick a specific non-reflexive Banach space and try specifying $f_n$? Maybe $X$ could be $c_0$?
– Junnan
Nov 18 at 15:41
add a comment |
2 Answers
2
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 $mathbb{N}timesmathbb{N}$.
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
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.
add a comment |
2 Answers
2
active
oldest
votes
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 $mathbb{N}timesmathbb{N}$.
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
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 $mathbb{N}timesmathbb{N}$.
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
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 $mathbb{N}timesmathbb{N}$.
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 $mathbb{N}timesmathbb{N}$.
edited Nov 18 at 22:29
answered Nov 18 at 20:04
Marco
1909
1909
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
add a comment |
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
There's one problem in your example which is K(I mean k with overbar, I don't know how to type it in comment) depends on n. So the first part of the sum in (2) cannot be controlled. Actually those two summable conditions are equivalent if by "summable" we mean "absolutely summable".
– Greywhite
2 days ago
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.
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.
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.
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 Nov 19 at 13:55
Greywhite
656
656
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998163%2fcomparing-two-summable-conditions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
Nov 18 at 15:41