Mixing
Norm in Sequence Space such that Convergence in Norm does not imply Pointwise Convergence
$begingroup$
I'm wandering if there exist a norm defined on the sequence space $mathbb{F}^{omega}$ such that there exist a sequence that converges in norm but it doesn't converge pointwise, element by element.
The same thing is asking for a sequence whose norm converges to $0$ but whose elements does not converge to $0$:
$$exists {x^{(n)}}, k mid lVert x^{(n)} lVert rightarrow 0, x^{(n)}_k nrightarrow 0$$
I couldn't find any example.
functional-analysis convergence normed-spaces
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
$endgroup$
|
show 1 more comment
$begingroup$
I'm wandering if there exist a norm defined on the sequence space $mathbb{F}^{omega}$ such that there exist a sequence that converges in norm but it doesn't converge pointwise, element by element.
The same thing is asking for a sequence whose norm converges to $0$ but whose elements does not converge to $0$:
$$exists {x^{(n)}}, k mid lVert x^{(n)} lVert rightarrow 0, x^{(n)}_k nrightarrow 0$$
I couldn't find any example.
functional-analysis convergence normed-spaces
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
$endgroup$
$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
1
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53
|
show 1 more comment
$begingroup$
I'm wandering if there exist a norm defined on the sequence space $mathbb{F}^{omega}$ such that there exist a sequence that converges in norm but it doesn't converge pointwise, element by element.
The same thing is asking for a sequence whose norm converges to $0$ but whose elements does not converge to $0$:
$$exists {x^{(n)}}, k mid lVert x^{(n)} lVert rightarrow 0, x^{(n)}_k nrightarrow 0$$
I couldn't find any example.
functional-analysis convergence normed-spaces
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
$endgroup$
I'm wandering if there exist a norm defined on the sequence space $mathbb{F}^{omega}$ such that there exist a sequence that converges in norm but it doesn't converge pointwise, element by element.
The same thing is asking for a sequence whose norm converges to $0$ but whose elements does not converge to $0$:
$$exists {x^{(n)}}, k mid lVert x^{(n)} lVert rightarrow 0, x^{(n)}_k nrightarrow 0$$
I couldn't find any example.
functional-analysis convergence normed-spaces
functional-analysis convergence normed-spaces
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
edited Feb 2 at 14:40
David C. Ullrich
61.7k44095
61.7k44095
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
asked Feb 2 at 11:55
Lorenzo NotaroLorenzo Notaro
384
384
$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
1
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53
|
show 1 more comment
$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
1
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53
$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
1
1
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$Lambda_j x=x_jquad(x=(x_1,x_2,dots))$$and note that what we need to prove is that there exists $j$ such that $Lambda_j$ is not bounded. Suppose each $Lambda_j$ is bounded, and define $x=(x_1,dots)$ by $$x_j=j||Lambda_j||.$$Then $$Lambda_jx=j||Lambda_j||,$$hence $$||x||ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||sum_{j=1}^nalpha_jb_j||=sum|alpha_j|$ for any $b_1,dots b_nin B$ and scalars $alpha_1,dots,alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $ell_infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=ell_infty$. Say $T:Xto X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $ell_infty$.)
Define $$||x||_X=||Tx||_infty.$$
Define $$Lambda_jx=x_jquad(x=(x_1,x_2,dots)).$$
We're done if there exists $j$ such that $Lambda_j$ is not bounded (with respect to $||cdot||_X$). Suppose to the contrary that every $Lambda_j$ is bounded. Now, for every $xin X$ we have $$sup_j|Lambda _j x|=||x||_infty<infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||Lambda_j||$ is bounded; say $||Lambda_j||le C$ for every $j$. This says precisely that $$||x||_inftyle C||x||_X=C||Tx||_infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
add a comment |
$begingroup$
The key thing to understand is that $mathbb{F}^omega$ is just some vector space of dimension $2^{aleph_0}$. As such, it is isomorphic to any other vector space of dimension $2^{aleph_0}$, which includes (for instance) any separable infinite-dimensional Banach space whatsoever. So, to construct a norm on $mathbb{F}^omega$, we don't have to explicitly write down a norm on sequences; we can just write down a norm on a totally different-looking vector space which happens to be isomorphic.
So, for instance, let $(x^{(n)})$ be any sequence of linearly independent elements of $mathbb{F}^omega$ which do not converge to $0$ pointwise. Pick your favorite separable infinite-dimensional Banach space $X$ with a sequence of linearly independent elements $(y^{(n)})$ which converge to $0$. Then there exists a vector space isomorphism $f:mathbb{F}^omegato X$ which maps $x^{(n)}$ to $y^{(n)}$ for each $n$. Transporting the norm of $X$ along $f$, we get a norm on $mathbb{F}^{omega}$ such that $x^{(n)}to 0$.
Alternatively, we could pick a Hamel basis for $mathbb{F}^omega$ which contains each $x^{(n)}$, and then define the norm explicitly in terms of the Hamel basis (for instance, we could use the $ell^infty$ norm for that basis, except with the $x^{(n)}$ coordinates scaled so that $x^{(n)}to 0$). This is again the same idea: a Hamel basis gives us an isomorphism $mathbb{F}^omegacongbigoplus_Imathbb{F}$ for some set $I$, and a norm is very easy to define on $bigoplus_Imathbb{F}$ since every element has only finitely many nonzero coordinates (so for instance, any of the usual $ell^p$ sequence norms are always finite).
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
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$begingroup$
Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$Lambda_j x=x_jquad(x=(x_1,x_2,dots))$$and note that what we need to prove is that there exists $j$ such that $Lambda_j$ is not bounded. Suppose each $Lambda_j$ is bounded, and define $x=(x_1,dots)$ by $$x_j=j||Lambda_j||.$$Then $$Lambda_jx=j||Lambda_j||,$$hence $$||x||ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||sum_{j=1}^nalpha_jb_j||=sum|alpha_j|$ for any $b_1,dots b_nin B$ and scalars $alpha_1,dots,alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $ell_infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=ell_infty$. Say $T:Xto X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $ell_infty$.)
Define $$||x||_X=||Tx||_infty.$$
Define $$Lambda_jx=x_jquad(x=(x_1,x_2,dots)).$$
We're done if there exists $j$ such that $Lambda_j$ is not bounded (with respect to $||cdot||_X$). Suppose to the contrary that every $Lambda_j$ is bounded. Now, for every $xin X$ we have $$sup_j|Lambda _j x|=||x||_infty<infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||Lambda_j||$ is bounded; say $||Lambda_j||le C$ for every $j$. This says precisely that $$||x||_inftyle C||x||_X=C||Tx||_infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
add a comment |
$begingroup$
Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$Lambda_j x=x_jquad(x=(x_1,x_2,dots))$$and note that what we need to prove is that there exists $j$ such that $Lambda_j$ is not bounded. Suppose each $Lambda_j$ is bounded, and define $x=(x_1,dots)$ by $$x_j=j||Lambda_j||.$$Then $$Lambda_jx=j||Lambda_j||,$$hence $$||x||ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||sum_{j=1}^nalpha_jb_j||=sum|alpha_j|$ for any $b_1,dots b_nin B$ and scalars $alpha_1,dots,alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $ell_infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=ell_infty$. Say $T:Xto X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $ell_infty$.)
Define $$||x||_X=||Tx||_infty.$$
Define $$Lambda_jx=x_jquad(x=(x_1,x_2,dots)).$$
We're done if there exists $j$ such that $Lambda_j$ is not bounded (with respect to $||cdot||_X$). Suppose to the contrary that every $Lambda_j$ is bounded. Now, for every $xin X$ we have $$sup_j|Lambda _j x|=||x||_infty<infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||Lambda_j||$ is bounded; say $||Lambda_j||le C$ for every $j$. This says precisely that $$||x||_inftyle C||x||_X=C||Tx||_infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
add a comment |
$begingroup$
Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$Lambda_j x=x_jquad(x=(x_1,x_2,dots))$$and note that what we need to prove is that there exists $j$ such that $Lambda_j$ is not bounded. Suppose each $Lambda_j$ is bounded, and define $x=(x_1,dots)$ by $$x_j=j||Lambda_j||.$$Then $$Lambda_jx=j||Lambda_j||,$$hence $$||x||ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||sum_{j=1}^nalpha_jb_j||=sum|alpha_j|$ for any $b_1,dots b_nin B$ and scalars $alpha_1,dots,alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $ell_infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=ell_infty$. Say $T:Xto X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $ell_infty$.)
Define $$||x||_X=||Tx||_infty.$$
Define $$Lambda_jx=x_jquad(x=(x_1,x_2,dots)).$$
We're done if there exists $j$ such that $Lambda_j$ is not bounded (with respect to $||cdot||_X$). Suppose to the contrary that every $Lambda_j$ is bounded. Now, for every $xin X$ we have $$sup_j|Lambda _j x|=||x||_infty<infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||Lambda_j||$ is bounded; say $||Lambda_j||le C$ for every $j$. This says precisely that $$||x||_inftyle C||x||_X=C||Tx||_infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$Lambda_j x=x_jquad(x=(x_1,x_2,dots))$$and note that what we need to prove is that there exists $j$ such that $Lambda_j$ is not bounded. Suppose each $Lambda_j$ is bounded, and define $x=(x_1,dots)$ by $$x_j=j||Lambda_j||.$$Then $$Lambda_jx=j||Lambda_j||,$$hence $$||x||ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||sum_{j=1}^nalpha_jb_j||=sum|alpha_j|$ for any $b_1,dots b_nin B$ and scalars $alpha_1,dots,alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $ell_infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=ell_infty$. Say $T:Xto X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $ell_infty$.)
Define $$||x||_X=||Tx||_infty.$$
Define $$Lambda_jx=x_jquad(x=(x_1,x_2,dots)).$$
We're done if there exists $j$ such that $Lambda_j$ is not bounded (with respect to $||cdot||_X$). Suppose to the contrary that every $Lambda_j$ is bounded. Now, for every $xin X$ we have $$sup_j|Lambda _j x|=||x||_infty<infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||Lambda_j||$ is bounded; say $||Lambda_j||le C$ for every $j$. This says precisely that $$||x||_inftyle C||x||_X=C||Tx||_infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
edited Feb 3 at 1:55
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 2 at 14:01
David C. UllrichDavid C. Ullrich
61.7k44095
61.7k44095
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
add a comment |
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
1
1
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
The question is about the space $mathbb{F}^omega$ of all sequences, not $ell^infty$ (of course, you can take a norm on $ell^infty$ and extend it to the full sequence space, but this may not be obvious to the asker and requires another use of AC). For what it's worth, I would be surprised if any norm can be defined on $mathbb{F}^omega$ without AC.
$endgroup$
– Eric Wofsey
Feb 2 at 16:56
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
$begingroup$
@EricWofsey Indeed, maybe the reason I misread the question is that the idea of treating $Bbb F^omega$ as a normed space seems curious. Turns out the question for $Bbb F^omega$ is completely and totally trivial.
$endgroup$
– David C. Ullrich
Feb 2 at 18:14
add a comment |
$begingroup$
The key thing to understand is that $mathbb{F}^omega$ is just some vector space of dimension $2^{aleph_0}$. As such, it is isomorphic to any other vector space of dimension $2^{aleph_0}$, which includes (for instance) any separable infinite-dimensional Banach space whatsoever. So, to construct a norm on $mathbb{F}^omega$, we don't have to explicitly write down a norm on sequences; we can just write down a norm on a totally different-looking vector space which happens to be isomorphic.
So, for instance, let $(x^{(n)})$ be any sequence of linearly independent elements of $mathbb{F}^omega$ which do not converge to $0$ pointwise. Pick your favorite separable infinite-dimensional Banach space $X$ with a sequence of linearly independent elements $(y^{(n)})$ which converge to $0$. Then there exists a vector space isomorphism $f:mathbb{F}^omegato X$ which maps $x^{(n)}$ to $y^{(n)}$ for each $n$. Transporting the norm of $X$ along $f$, we get a norm on $mathbb{F}^{omega}$ such that $x^{(n)}to 0$.
Alternatively, we could pick a Hamel basis for $mathbb{F}^omega$ which contains each $x^{(n)}$, and then define the norm explicitly in terms of the Hamel basis (for instance, we could use the $ell^infty$ norm for that basis, except with the $x^{(n)}$ coordinates scaled so that $x^{(n)}to 0$). This is again the same idea: a Hamel basis gives us an isomorphism $mathbb{F}^omegacongbigoplus_Imathbb{F}$ for some set $I$, and a norm is very easy to define on $bigoplus_Imathbb{F}$ since every element has only finitely many nonzero coordinates (so for instance, any of the usual $ell^p$ sequence norms are always finite).
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
The key thing to understand is that $mathbb{F}^omega$ is just some vector space of dimension $2^{aleph_0}$. As such, it is isomorphic to any other vector space of dimension $2^{aleph_0}$, which includes (for instance) any separable infinite-dimensional Banach space whatsoever. So, to construct a norm on $mathbb{F}^omega$, we don't have to explicitly write down a norm on sequences; we can just write down a norm on a totally different-looking vector space which happens to be isomorphic.
So, for instance, let $(x^{(n)})$ be any sequence of linearly independent elements of $mathbb{F}^omega$ which do not converge to $0$ pointwise. Pick your favorite separable infinite-dimensional Banach space $X$ with a sequence of linearly independent elements $(y^{(n)})$ which converge to $0$. Then there exists a vector space isomorphism $f:mathbb{F}^omegato X$ which maps $x^{(n)}$ to $y^{(n)}$ for each $n$. Transporting the norm of $X$ along $f$, we get a norm on $mathbb{F}^{omega}$ such that $x^{(n)}to 0$.
Alternatively, we could pick a Hamel basis for $mathbb{F}^omega$ which contains each $x^{(n)}$, and then define the norm explicitly in terms of the Hamel basis (for instance, we could use the $ell^infty$ norm for that basis, except with the $x^{(n)}$ coordinates scaled so that $x^{(n)}to 0$). This is again the same idea: a Hamel basis gives us an isomorphism $mathbb{F}^omegacongbigoplus_Imathbb{F}$ for some set $I$, and a norm is very easy to define on $bigoplus_Imathbb{F}$ since every element has only finitely many nonzero coordinates (so for instance, any of the usual $ell^p$ sequence norms are always finite).
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
$endgroup$
add a comment |
$begingroup$
The key thing to understand is that $mathbb{F}^omega$ is just some vector space of dimension $2^{aleph_0}$. As such, it is isomorphic to any other vector space of dimension $2^{aleph_0}$, which includes (for instance) any separable infinite-dimensional Banach space whatsoever. So, to construct a norm on $mathbb{F}^omega$, we don't have to explicitly write down a norm on sequences; we can just write down a norm on a totally different-looking vector space which happens to be isomorphic.
So, for instance, let $(x^{(n)})$ be any sequence of linearly independent elements of $mathbb{F}^omega$ which do not converge to $0$ pointwise. Pick your favorite separable infinite-dimensional Banach space $X$ with a sequence of linearly independent elements $(y^{(n)})$ which converge to $0$. Then there exists a vector space isomorphism $f:mathbb{F}^omegato X$ which maps $x^{(n)}$ to $y^{(n)}$ for each $n$. Transporting the norm of $X$ along $f$, we get a norm on $mathbb{F}^{omega}$ such that $x^{(n)}to 0$.
Alternatively, we could pick a Hamel basis for $mathbb{F}^omega$ which contains each $x^{(n)}$, and then define the norm explicitly in terms of the Hamel basis (for instance, we could use the $ell^infty$ norm for that basis, except with the $x^{(n)}$ coordinates scaled so that $x^{(n)}to 0$). This is again the same idea: a Hamel basis gives us an isomorphism $mathbb{F}^omegacongbigoplus_Imathbb{F}$ for some set $I$, and a norm is very easy to define on $bigoplus_Imathbb{F}$ since every element has only finitely many nonzero coordinates (so for instance, any of the usual $ell^p$ sequence norms are always finite).
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
$endgroup$
The key thing to understand is that $mathbb{F}^omega$ is just some vector space of dimension $2^{aleph_0}$. As such, it is isomorphic to any other vector space of dimension $2^{aleph_0}$, which includes (for instance) any separable infinite-dimensional Banach space whatsoever. So, to construct a norm on $mathbb{F}^omega$, we don't have to explicitly write down a norm on sequences; we can just write down a norm on a totally different-looking vector space which happens to be isomorphic.
So, for instance, let $(x^{(n)})$ be any sequence of linearly independent elements of $mathbb{F}^omega$ which do not converge to $0$ pointwise. Pick your favorite separable infinite-dimensional Banach space $X$ with a sequence of linearly independent elements $(y^{(n)})$ which converge to $0$. Then there exists a vector space isomorphism $f:mathbb{F}^omegato X$ which maps $x^{(n)}$ to $y^{(n)}$ for each $n$. Transporting the norm of $X$ along $f$, we get a norm on $mathbb{F}^{omega}$ such that $x^{(n)}to 0$.
Alternatively, we could pick a Hamel basis for $mathbb{F}^omega$ which contains each $x^{(n)}$, and then define the norm explicitly in terms of the Hamel basis (for instance, we could use the $ell^infty$ norm for that basis, except with the $x^{(n)}$ coordinates scaled so that $x^{(n)}to 0$). This is again the same idea: a Hamel basis gives us an isomorphism $mathbb{F}^omegacongbigoplus_Imathbb{F}$ for some set $I$, and a norm is very easy to define on $bigoplus_Imathbb{F}$ since every element has only finitely many nonzero coordinates (so for instance, any of the usual $ell^p$ sequence norms are always finite).
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 16:52
Eric WofseyEric Wofsey
193k14221352
193k14221352
add a comment |
add a comment |
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$begingroup$
@David: I'm not sure this is really about using choice. If we start tagging every question which admits a solution using choice under the tag, where does it end?
$endgroup$
– Asaf Karagila♦
Feb 2 at 14:27
$begingroup$
@AsafKaragila Fine. I've removed the tag, since the only reason I added it was to try to attract the attention of someone who might know the answer to a question implicit in my answer; it's done that. Q: Is it consistent with ZF that every linear functional on a Banach space is bounded? (One has the general impression that "there are no interesting counterexamples in real analysis" is consistent with ZF...)
$endgroup$
– David C. Ullrich
Feb 2 at 14:42
$begingroup$
@David: It is, yes. It follows from automatic continuity assumptions, e.g. "Every set of reals is Lebesgue measurable" or "Every set of reals has the Baire Property".
$endgroup$
– Asaf Karagila♦
Feb 2 at 15:40
$begingroup$
@AsafKaragila Of course those automatic continuity things were why I thought it might be so. I don't see offhand how they actually imply that every linear functional on a Banach space is bounded - is it possible to give a clue in a comment?
$endgroup$
– David C. Ullrich
Feb 2 at 17:43
1
$begingroup$
@David: It's not trivial. You can find details in my note "Zornian Functional Analysis" (which you can find on my home page (which you can find through my profile)). If my memory serves me right, it's simpler for the separable case, and far more intricate in the general case. (But I might be confusing this with a different theorem of this flavor...)
$endgroup$
– Asaf Karagila♦
Feb 2 at 18:53