Show that the image of $T:l^inftyto l^infty$, $(x_n)_n mapsto Big(fracx_nnBig)_n$ is not closed in $l^infty$. The 2019 Stack Overflow Developer Survey Results Are InShow that operator is continuous and selfadjoint (or not)show that the operator $T:l^2rightarrow V$ is boundedIs the norm on $ell^infty$ induced by an inner product?Show that $c_0$ is a Banach space with the norm $rVert cdot lVert_infty$Show that $ x_n oversetTmapsto sum_k=1^infty a_nk x_k $ is compactBounded and surjective map from counting $L^1$ to separable Banach space.Real Analysis, Folland problem 5.5.36Problem about a compact operator $T:l^prightarrow l^p$Prove $C^1([0,1]) $ is Banach.Show that the inverse function of the identity $textidcolon (C[0,1],lVertcdotrVert_infty)to (C[0,1],lVertcdotrVert_1)$ is not continuous
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Show that the image of $T:l^inftyto l^infty$, $(x_n)_n mapsto Big(fracx_nnBig)_n$ is not closed in $l^infty$.
The 2019 Stack Overflow Developer Survey Results Are InShow that operator is continuous and selfadjoint (or not)show that the operator $T:l^2rightarrow V$ is boundedIs the norm on $ell^infty$ induced by an inner product?Show that $c_0$ is a Banach space with the norm $rVert cdot lVert_infty$Show that $ x_n oversetTmapsto sum_k=1^infty a_nk x_k $ is compactBounded and surjective map from counting $L^1$ to separable Banach space.Real Analysis, Folland problem 5.5.36Problem about a compact operator $T:l^prightarrow l^p$Prove $C^1([0,1]) $ is Banach.Show that the inverse function of the identity $textidcolon (C[0,1],lVertcdotrVert_infty)to (C[0,1],lVertcdotrVert_1)$ is not continuous
$begingroup$
Denote the set of all bounded sequences in $mathbbR$ by $l^infty$, endowed with the sup norm $lVert rVert _infty$. Define a map $T:l^infty to l^infty$ as follows:
$$(x_n)_n mapsto Big(fracx_nnBig)_n.$$
Then the question is:
Show that the image of $l^infty$ under $T$ is not a closed set in $l^infty$.
My attempts:
As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^infty)$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, ldots, 1, ldots) = t_0$$ $(notin T(l^infty))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^infty)$ which goes to it. The following are two (failed) attempts to construct such a sequence.
First I thought the sequences $$t_n = (1,1,ldots,1,frac1(n+1)^2,frac1(n+2)^2,ldots)$$ would work. But they didn't, because
$$lVert t_n-t_0rVert_infty = sup_kgeq nBig|1-frac1k^2Big|=1 nrightarrow 0$$ as $nto infty$ and thus $(t_n) nrightarrow t_0$.
Also I thought about the sequences $$t_n = (1,1,ldots,1,1+frac1n+1,1+frac1n+2,ldots,1+frac1n+m,ldots).$$ In this case $(t_n) to t_0$ as $n to infty$, but none of the $t_n$'s is an element of $T(l^infty)$. So this sequence also is of no use.
Any help? Thanks a lot in advance.
sequences-and-series functional-analysis banach-spaces
edited Apr 21 '13 at 3:45
Julien
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
$endgroup$
|
show 4 more comments
$begingroup$
Denote the set of all bounded sequences in $mathbbR$ by $l^infty$, endowed with the sup norm $lVert rVert _infty$. Define a map $T:l^infty to l^infty$ as follows:
$$(x_n)_n mapsto Big(fracx_nnBig)_n.$$
Then the question is:
Show that the image of $l^infty$ under $T$ is not a closed set in $l^infty$.
My attempts:
As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^infty)$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, ldots, 1, ldots) = t_0$$ $(notin T(l^infty))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^infty)$ which goes to it. The following are two (failed) attempts to construct such a sequence.
First I thought the sequences $$t_n = (1,1,ldots,1,frac1(n+1)^2,frac1(n+2)^2,ldots)$$ would work. But they didn't, because
$$lVert t_n-t_0rVert_infty = sup_kgeq nBig|1-frac1k^2Big|=1 nrightarrow 0$$ as $nto infty$ and thus $(t_n) nrightarrow t_0$.
Also I thought about the sequences $$t_n = (1,1,ldots,1,1+frac1n+1,1+frac1n+2,ldots,1+frac1n+m,ldots).$$ In this case $(t_n) to t_0$ as $n to infty$, but none of the $t_n$'s is an element of $T(l^infty)$. So this sequence also is of no use.
Any help? Thanks a lot in advance.
sequences-and-series functional-analysis banach-spaces
edited Apr 21 '13 at 3:45
Julien
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
$endgroup$
$begingroup$
$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
$endgroup$
– Diego
Apr 21 '13 at 3:25
$begingroup$
Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
$endgroup$
– Suugaku
Apr 21 '13 at 3:27
2
$begingroup$
$1-fracmm+1<1-fracmm+2<...$
$endgroup$
– Diego
Apr 21 '13 at 3:37
1
$begingroup$
The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
$endgroup$
– Julien
Apr 21 '13 at 3:43
1
$begingroup$
Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
$endgroup$
– Julien
Apr 21 '13 at 3:53
|
show 4 more comments
$begingroup$
Denote the set of all bounded sequences in $mathbbR$ by $l^infty$, endowed with the sup norm $lVert rVert _infty$. Define a map $T:l^infty to l^infty$ as follows:
$$(x_n)_n mapsto Big(fracx_nnBig)_n.$$
Then the question is:
Show that the image of $l^infty$ under $T$ is not a closed set in $l^infty$.
My attempts:
As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^infty)$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, ldots, 1, ldots) = t_0$$ $(notin T(l^infty))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^infty)$ which goes to it. The following are two (failed) attempts to construct such a sequence.
First I thought the sequences $$t_n = (1,1,ldots,1,frac1(n+1)^2,frac1(n+2)^2,ldots)$$ would work. But they didn't, because
$$lVert t_n-t_0rVert_infty = sup_kgeq nBig|1-frac1k^2Big|=1 nrightarrow 0$$ as $nto infty$ and thus $(t_n) nrightarrow t_0$.
Also I thought about the sequences $$t_n = (1,1,ldots,1,1+frac1n+1,1+frac1n+2,ldots,1+frac1n+m,ldots).$$ In this case $(t_n) to t_0$ as $n to infty$, but none of the $t_n$'s is an element of $T(l^infty)$. So this sequence also is of no use.
Any help? Thanks a lot in advance.
sequences-and-series functional-analysis banach-spaces
edited Apr 21 '13 at 3:45
Julien
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
$endgroup$
Denote the set of all bounded sequences in $mathbbR$ by $l^infty$, endowed with the sup norm $lVert rVert _infty$. Define a map $T:l^infty to l^infty$ as follows:
$$(x_n)_n mapsto Big(fracx_nnBig)_n.$$
Then the question is:
Show that the image of $l^infty$ under $T$ is not a closed set in $l^infty$.
My attempts:
As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^infty)$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, ldots, 1, ldots) = t_0$$ $(notin T(l^infty))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^infty)$ which goes to it. The following are two (failed) attempts to construct such a sequence.
First I thought the sequences $$t_n = (1,1,ldots,1,frac1(n+1)^2,frac1(n+2)^2,ldots)$$ would work. But they didn't, because
$$lVert t_n-t_0rVert_infty = sup_kgeq nBig|1-frac1k^2Big|=1 nrightarrow 0$$ as $nto infty$ and thus $(t_n) nrightarrow t_0$.
Also I thought about the sequences $$t_n = (1,1,ldots,1,1+frac1n+1,1+frac1n+2,ldots,1+frac1n+m,ldots).$$ In this case $(t_n) to t_0$ as $n to infty$, but none of the $t_n$'s is an element of $T(l^infty)$. So this sequence also is of no use.
Any help? Thanks a lot in advance.
sequences-and-series functional-analysis banach-spaces
sequences-and-series functional-analysis banach-spaces
edited Apr 21 '13 at 3:45
Julien
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
edited Apr 21 '13 at 3:45
Julien
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
edited Apr 21 '13 at 3:45
Julien
38.8k358132
edited Apr 21 '13 at 3:45
Julien
38.8k358132
edited Apr 21 '13 at 3:45
Julien
38.8k358132
38.8k358132
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
asked Apr 21 '13 at 2:53
SayantanSayantan
1,54321423
1,54321423
$begingroup$
$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
$endgroup$
– Diego
Apr 21 '13 at 3:25
$begingroup$
Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
$endgroup$
– Suugaku
Apr 21 '13 at 3:27
2
$begingroup$
$1-fracmm+1<1-fracmm+2<...$
$endgroup$
– Diego
Apr 21 '13 at 3:37
1
$begingroup$
The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
$endgroup$
– Julien
Apr 21 '13 at 3:43
1
$begingroup$
Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
$endgroup$
– Julien
Apr 21 '13 at 3:53
|
show 4 more comments
$begingroup$
$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
$endgroup$
– Diego
Apr 21 '13 at 3:25
$begingroup$
Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
$endgroup$
– Suugaku
Apr 21 '13 at 3:27
2
$begingroup$
$1-fracmm+1<1-fracmm+2<...$
$endgroup$
– Diego
Apr 21 '13 at 3:37
1
$begingroup$
The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
$endgroup$
– Julien
Apr 21 '13 at 3:43
1
$begingroup$
Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
$endgroup$
– Julien
Apr 21 '13 at 3:53
$begingroup$
$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
$endgroup$
– Diego
Apr 21 '13 at 3:25
$begingroup$
$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
$endgroup$
– Diego
Apr 21 '13 at 3:25
$begingroup$
Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
$endgroup$
– Suugaku
Apr 21 '13 at 3:27
$begingroup$
Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
$endgroup$
– Suugaku
Apr 21 '13 at 3:27
2
2
$begingroup$
$1-fracmm+1<1-fracmm+2<...$
$endgroup$
– Diego
Apr 21 '13 at 3:37
$begingroup$
$1-fracmm+1<1-fracmm+2<...$
$endgroup$
– Diego
Apr 21 '13 at 3:37
1
1
$begingroup$
The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
$endgroup$
– Julien
Apr 21 '13 at 3:43
$begingroup$
The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
$endgroup$
– Julien
Apr 21 '13 at 3:43
1
1
$begingroup$
Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
$endgroup$
– Julien
Apr 21 '13 at 3:53
$begingroup$
Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
$endgroup$
– Julien
Apr 21 '13 at 3:53
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $t_0=(1,frac1sqrt2,frac1sqrt3,ldots)$ and $t_m=(1,frac1sqrt2,ldots,frac1sqrtm,0,0,ldots)$. Clearly $t_mrightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
$endgroup$
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
add a comment |
$begingroup$
Consider $t_0=(frac 1sqrtn)_n$. This too is not in the image of $T$, since its preimage is $(sqrtn)_n$. Now define $t_m=(frac1sqrt1,frac1sqrt2,ldots,frac1sqrtm,fracsqrtmm+1,fracsqrtmm+2,ldots)$, the image of $(sqrt1,sqrt2,ldots,sqrtm,sqrtm,sqrtm,ldots)$. But now $|t_m-t_0|_infty=sup_n> m=sup_n> m<frac1m$, because $0<|sqrtfracmn-1|<1$
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
$endgroup$
$begingroup$
Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $t_0=(1,frac1sqrt2,frac1sqrt3,ldots)$ and $t_m=(1,frac1sqrt2,ldots,frac1sqrtm,0,0,ldots)$. Clearly $t_mrightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
$endgroup$
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
add a comment |
$begingroup$
Let $t_0=(1,frac1sqrt2,frac1sqrt3,ldots)$ and $t_m=(1,frac1sqrt2,ldots,frac1sqrtm,0,0,ldots)$. Clearly $t_mrightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
$endgroup$
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
add a comment |
$begingroup$
Let $t_0=(1,frac1sqrt2,frac1sqrt3,ldots)$ and $t_m=(1,frac1sqrt2,ldots,frac1sqrtm,0,0,ldots)$. Clearly $t_mrightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
$endgroup$
Let $t_0=(1,frac1sqrt2,frac1sqrt3,ldots)$ and $t_m=(1,frac1sqrt2,ldots,frac1sqrtm,0,0,ldots)$. Clearly $t_mrightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
answered Apr 21 '13 at 3:35
DiegoDiego
2,007716
2,007716
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
add a comment |
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Oh, that's much better than my example, +1.
$endgroup$
– Julien
Apr 21 '13 at 3:37
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
Thanks for the really nice example, +1. :-)
$endgroup$
– Sayantan
Apr 21 '13 at 3:46
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
$begingroup$
You are welcome! :)
$endgroup$
– Diego
Apr 21 '13 at 3:52
add a comment |
$begingroup$
Consider $t_0=(frac 1sqrtn)_n$. This too is not in the image of $T$, since its preimage is $(sqrtn)_n$. Now define $t_m=(frac1sqrt1,frac1sqrt2,ldots,frac1sqrtm,fracsqrtmm+1,fracsqrtmm+2,ldots)$, the image of $(sqrt1,sqrt2,ldots,sqrtm,sqrtm,sqrtm,ldots)$. But now $|t_m-t_0|_infty=sup_n> m=sup_n> m<frac1m$, because $0<|sqrtfracmn-1|<1$
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
$endgroup$
$begingroup$
Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
add a comment |
$begingroup$
Consider $t_0=(frac 1sqrtn)_n$. This too is not in the image of $T$, since its preimage is $(sqrtn)_n$. Now define $t_m=(frac1sqrt1,frac1sqrt2,ldots,frac1sqrtm,fracsqrtmm+1,fracsqrtmm+2,ldots)$, the image of $(sqrt1,sqrt2,ldots,sqrtm,sqrtm,sqrtm,ldots)$. But now $|t_m-t_0|_infty=sup_n> m=sup_n> m<frac1m$, because $0<|sqrtfracmn-1|<1$
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
$endgroup$
$begingroup$
Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
add a comment |
$begingroup$
Consider $t_0=(frac 1sqrtn)_n$. This too is not in the image of $T$, since its preimage is $(sqrtn)_n$. Now define $t_m=(frac1sqrt1,frac1sqrt2,ldots,frac1sqrtm,fracsqrtmm+1,fracsqrtmm+2,ldots)$, the image of $(sqrt1,sqrt2,ldots,sqrtm,sqrtm,sqrtm,ldots)$. But now $|t_m-t_0|_infty=sup_n> m=sup_n> m<frac1m$, because $0<|sqrtfracmn-1|<1$
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
$endgroup$
Consider $t_0=(frac 1sqrtn)_n$. This too is not in the image of $T$, since its preimage is $(sqrtn)_n$. Now define $t_m=(frac1sqrt1,frac1sqrt2,ldots,frac1sqrtm,fracsqrtmm+1,fracsqrtmm+2,ldots)$, the image of $(sqrt1,sqrt2,ldots,sqrtm,sqrtm,sqrtm,ldots)$. But now $|t_m-t_0|_infty=sup_n> m=sup_n> m<frac1m$, because $0<|sqrtfracmn-1|<1$
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
edited Apr 21 '13 at 3:45
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
answered Apr 21 '13 at 3:06
vadim123vadim123
76.6k897191
76.6k897191
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Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
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– Amzoti
Apr 21 '13 at 3:34
add a comment |
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Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
$begingroup$
Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
$begingroup$
Answers should be constructive and should not add opinions that might make OPs feel degraded because they may not have understood something. I believe you got a downvote because of this and may want to update your answer. Regards
$endgroup$
– Amzoti
Apr 21 '13 at 3:34
add a comment |
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$t_0$ will not work because $B(t_0,1/2)$ has empty intersection with $T(l^infty)$.
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– Diego
Apr 21 '13 at 3:25
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Hi Diego, I'd like to know what did I do incorrectly below if your statement is the case? If it's wrong, I want to delete it!
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– Suugaku
Apr 21 '13 at 3:27
2
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$1-fracmm+1<1-fracmm+2<...$
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– Diego
Apr 21 '13 at 3:37
1
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The idea behind Diego's example is the following: the range of $T$ contains the finitely supported sequences and is contained in $c_0$. It follows that the closure of the range of $T$ is $c_0$. And not every $c_0$ sequence is in the range of $T$, so it is not closed.
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– Julien
Apr 21 '13 at 3:43
1
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Finitely supported: finitely many nonzero terms. $c_0$: sequences which tend to $0$. The former are dense in the latter. All this in $ell^infty$.
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– Julien
Apr 21 '13 at 3:53