$T$ has not a closed range
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Let $$T = {rm diag}(0, 1, 0, frac{1}{2!}, 0, frac{1}{3!}, dots)$$
Clearly $T$ is a positive operator on the Hilbert space $ell_{mathbb{N}^*}^2(mathbb{C})$.
I want to prove that $T$ has not a closed range.
functional-analysis operator-theory
asked Nov 12 at 17:44
Schüler
1,4631421
add a comment |
up vote
2
down vote
favorite
Let $$T = {rm diag}(0, 1, 0, frac{1}{2!}, 0, frac{1}{3!}, dots)$$
Clearly $T$ is a positive operator on the Hilbert space $ell_{mathbb{N}^*}^2(mathbb{C})$.
I want to prove that $T$ has not a closed range.
functional-analysis operator-theory
asked Nov 12 at 17:44
Schüler
1,4631421
what have you tried?
– supinf
Nov 12 at 17:51
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $$T = {rm diag}(0, 1, 0, frac{1}{2!}, 0, frac{1}{3!}, dots)$$
Clearly $T$ is a positive operator on the Hilbert space $ell_{mathbb{N}^*}^2(mathbb{C})$.
I want to prove that $T$ has not a closed range.
functional-analysis operator-theory
asked Nov 12 at 17:44
Schüler
1,4631421
Let $$T = {rm diag}(0, 1, 0, frac{1}{2!}, 0, frac{1}{3!}, dots)$$
Clearly $T$ is a positive operator on the Hilbert space $ell_{mathbb{N}^*}^2(mathbb{C})$.
I want to prove that $T$ has not a closed range.
functional-analysis operator-theory
functional-analysis operator-theory
asked Nov 12 at 17:44
Schüler
1,4631421
asked Nov 12 at 17:44
Schüler
1,4631421
asked Nov 12 at 17:44
Schüler
1,4631421
asked Nov 12 at 17:44
Schüler
1,4631421
asked Nov 12 at 17:44
Schüler
1,4631421
1,4631421
what have you tried?
– supinf
Nov 12 at 17:51
add a comment |
what have you tried?
– supinf
Nov 12 at 17:51
what have you tried?
– supinf
Nov 12 at 17:51
what have you tried?
– supinf
Nov 12 at 17:51
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
Write ${e_n}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $operatorname{ran}Tsubset X$, where $$X=overline{operatorname{span}}{e_{2n}: ninmathbb N}.$$
For any $n$, $Te_{2n}=tfrac1{n!},e_{2n}.$. So $$e_{2n}=T(n!,e_{2n})in operatorname{ran}T.$$ So $operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $operatorname{ran}T$. Consider for instance
$$
x=sum_{n=1}^infty tfrac1n,e_{2n}.
$$
To have $xinoperatorname{ran}T$, say $x=Ty$, where $y=sum_{n=1}^infty c_ne_{2n}$ and $sum_n|c_n|^2<infty$. From $x=Ty$ we get
$$
tfrac1n,e_{2n}=tfrac1{n!},c_n,e_{2n},
$$
which would require $c_n=(n-1)!$.
answered Nov 12 at 19:21
Martin Argerami
121k1072172
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Write ${e_n}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $operatorname{ran}Tsubset X$, where $$X=overline{operatorname{span}}{e_{2n}: ninmathbb N}.$$
For any $n$, $Te_{2n}=tfrac1{n!},e_{2n}.$. So $$e_{2n}=T(n!,e_{2n})in operatorname{ran}T.$$ So $operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $operatorname{ran}T$. Consider for instance
$$
x=sum_{n=1}^infty tfrac1n,e_{2n}.
$$
To have $xinoperatorname{ran}T$, say $x=Ty$, where $y=sum_{n=1}^infty c_ne_{2n}$ and $sum_n|c_n|^2<infty$. From $x=Ty$ we get
$$
tfrac1n,e_{2n}=tfrac1{n!},c_n,e_{2n},
$$
which would require $c_n=(n-1)!$.
answered Nov 12 at 19:21
Martin Argerami
121k1072172
add a comment |
up vote
3
down vote
Write ${e_n}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $operatorname{ran}Tsubset X$, where $$X=overline{operatorname{span}}{e_{2n}: ninmathbb N}.$$
For any $n$, $Te_{2n}=tfrac1{n!},e_{2n}.$. So $$e_{2n}=T(n!,e_{2n})in operatorname{ran}T.$$ So $operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $operatorname{ran}T$. Consider for instance
$$
x=sum_{n=1}^infty tfrac1n,e_{2n}.
$$
To have $xinoperatorname{ran}T$, say $x=Ty$, where $y=sum_{n=1}^infty c_ne_{2n}$ and $sum_n|c_n|^2<infty$. From $x=Ty$ we get
$$
tfrac1n,e_{2n}=tfrac1{n!},c_n,e_{2n},
$$
which would require $c_n=(n-1)!$.
answered Nov 12 at 19:21
Martin Argerami
121k1072172
add a comment |
up vote
3
down vote
up vote
3
down vote
Write ${e_n}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $operatorname{ran}Tsubset X$, where $$X=overline{operatorname{span}}{e_{2n}: ninmathbb N}.$$
For any $n$, $Te_{2n}=tfrac1{n!},e_{2n}.$. So $$e_{2n}=T(n!,e_{2n})in operatorname{ran}T.$$ So $operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $operatorname{ran}T$. Consider for instance
$$
x=sum_{n=1}^infty tfrac1n,e_{2n}.
$$
To have $xinoperatorname{ran}T$, say $x=Ty$, where $y=sum_{n=1}^infty c_ne_{2n}$ and $sum_n|c_n|^2<infty$. From $x=Ty$ we get
$$
tfrac1n,e_{2n}=tfrac1{n!},c_n,e_{2n},
$$
which would require $c_n=(n-1)!$.
answered Nov 12 at 19:21
Martin Argerami
121k1072172
Write ${e_n}$ for the canonical basis. You have that $Te_{2n-1}=0$ for all $n$, so $operatorname{ran}Tsubset X$, where $$X=overline{operatorname{span}}{e_{2n}: ninmathbb N}.$$
For any $n$, $Te_{2n}=tfrac1{n!},e_{2n}.$. So $$e_{2n}=T(n!,e_{2n})in operatorname{ran}T.$$ So $operatorname{ran}T$ is dense in $X$. But there are elements in $X$ that are no in $operatorname{ran}T$. Consider for instance
$$
x=sum_{n=1}^infty tfrac1n,e_{2n}.
$$
To have $xinoperatorname{ran}T$, say $x=Ty$, where $y=sum_{n=1}^infty c_ne_{2n}$ and $sum_n|c_n|^2<infty$. From $x=Ty$ we get
$$
tfrac1n,e_{2n}=tfrac1{n!},c_n,e_{2n},
$$
which would require $c_n=(n-1)!$.
answered Nov 12 at 19:21
Martin Argerami
121k1072172
answered Nov 12 at 19:21
Martin Argerami
121k1072172
answered Nov 12 at 19:21
Martin Argerami
121k1072172
answered Nov 12 at 19:21
Martin Argerami
121k1072172
121k1072172
add a comment |
add a comment |
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what have you tried?
– supinf
Nov 12 at 17:51