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If $r(a) < 1$, does $sum_n=0^infty a^*na^n$ necessarily converge?
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In Murphy, exercise 2.6:
Let $A$ be a unital C$^*$-algebra. If $r(a) < 1$ and $b = (sum_n=0^infty a^*na^n)^1/2$, show that $b geq 1$ and $lVert bab^-1rVert < 1$....
where $r(a)$ denotes the spectral radius of $a$. Is the convergence of $sum_n=0^infty a^*na^n$ an assumption of the problem, or does it follow from $r(a) < 1$? If it does not follow from $r(a) < 1$, is there a counter example? I can't seem to make any progress in either direction.
operator-algebras spectral-theory c-star-algebras
asked Jul 28 at 19:15
AGF
204
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In Murphy, exercise 2.6:
Let $A$ be a unital C$^*$-algebra. If $r(a) < 1$ and $b = (sum_n=0^infty a^*na^n)^1/2$, show that $b geq 1$ and $lVert bab^-1rVert < 1$....
where $r(a)$ denotes the spectral radius of $a$. Is the convergence of $sum_n=0^infty a^*na^n$ an assumption of the problem, or does it follow from $r(a) < 1$? If it does not follow from $r(a) < 1$, is there a counter example? I can't seem to make any progress in either direction.
operator-algebras spectral-theory c-star-algebras
asked Jul 28 at 19:15
AGF
204
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Murphy, exercise 2.6:
Let $A$ be a unital C$^*$-algebra. If $r(a) < 1$ and $b = (sum_n=0^infty a^*na^n)^1/2$, show that $b geq 1$ and $lVert bab^-1rVert < 1$....
where $r(a)$ denotes the spectral radius of $a$. Is the convergence of $sum_n=0^infty a^*na^n$ an assumption of the problem, or does it follow from $r(a) < 1$? If it does not follow from $r(a) < 1$, is there a counter example? I can't seem to make any progress in either direction.
operator-algebras spectral-theory c-star-algebras
asked Jul 28 at 19:15
AGF
204
In Murphy, exercise 2.6:
Let $A$ be a unital C$^*$-algebra. If $r(a) < 1$ and $b = (sum_n=0^infty a^*na^n)^1/2$, show that $b geq 1$ and $lVert bab^-1rVert < 1$....
where $r(a)$ denotes the spectral radius of $a$. Is the convergence of $sum_n=0^infty a^*na^n$ an assumption of the problem, or does it follow from $r(a) < 1$? If it does not follow from $r(a) < 1$, is there a counter example? I can't seem to make any progress in either direction.
operator-algebras spectral-theory c-star-algebras
asked Jul 28 at 19:15
AGF
204
edited Jul 28 at 19:43
asked Jul 28 at 19:15
AGF
204
asked Jul 28 at 19:15
AGF
204
asked Jul 28 at 19:15
AGF
204
204
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1 Answer
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accepted
Since $r(a)<1$, from
$$
r(a)=lim_n|a^n|^1/n=lim_n|a^*na^n|^1/2n
$$
it follows that there exist $c<1$ such that for all $n$ big enough, $$|a^*na^n|leq c^2n.$$ So the series always converges.
For a little more detail, given $varepsilon>0$ with $c=r(a)+varepsilon < 1$, there exists $n_0$ such that for all $ngeq n_0$ we have $$ |r(a)-|a^*na^n|^1/2n|<varepsilon.$$ In particular,
$$
|a^*na^n|^1/2n<r(a)+varepsilon = c,
$$
which gives
$$
|a^*na^n|leq c^2n.
$$
answered Jul 28 at 19:58
Martin Argerami
115k1071164
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $r(a)<1$, from
$$
r(a)=lim_n|a^n|^1/n=lim_n|a^*na^n|^1/2n
$$
it follows that there exist $c<1$ such that for all $n$ big enough, $$|a^*na^n|leq c^2n.$$ So the series always converges.
For a little more detail, given $varepsilon>0$ with $c=r(a)+varepsilon < 1$, there exists $n_0$ such that for all $ngeq n_0$ we have $$ |r(a)-|a^*na^n|^1/2n|<varepsilon.$$ In particular,
$$
|a^*na^n|^1/2n<r(a)+varepsilon = c,
$$
which gives
$$
|a^*na^n|leq c^2n.
$$
answered Jul 28 at 19:58
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
Since $r(a)<1$, from
$$
r(a)=lim_n|a^n|^1/n=lim_n|a^*na^n|^1/2n
$$
it follows that there exist $c<1$ such that for all $n$ big enough, $$|a^*na^n|leq c^2n.$$ So the series always converges.
For a little more detail, given $varepsilon>0$ with $c=r(a)+varepsilon < 1$, there exists $n_0$ such that for all $ngeq n_0$ we have $$ |r(a)-|a^*na^n|^1/2n|<varepsilon.$$ In particular,
$$
|a^*na^n|^1/2n<r(a)+varepsilon = c,
$$
which gives
$$
|a^*na^n|leq c^2n.
$$
answered Jul 28 at 19:58
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $r(a)<1$, from
$$
r(a)=lim_n|a^n|^1/n=lim_n|a^*na^n|^1/2n
$$
it follows that there exist $c<1$ such that for all $n$ big enough, $$|a^*na^n|leq c^2n.$$ So the series always converges.
For a little more detail, given $varepsilon>0$ with $c=r(a)+varepsilon < 1$, there exists $n_0$ such that for all $ngeq n_0$ we have $$ |r(a)-|a^*na^n|^1/2n|<varepsilon.$$ In particular,
$$
|a^*na^n|^1/2n<r(a)+varepsilon = c,
$$
which gives
$$
|a^*na^n|leq c^2n.
$$
answered Jul 28 at 19:58
Martin Argerami
115k1071164
Since $r(a)<1$, from
$$
r(a)=lim_n|a^n|^1/n=lim_n|a^*na^n|^1/2n
$$
it follows that there exist $c<1$ such that for all $n$ big enough, $$|a^*na^n|leq c^2n.$$ So the series always converges.
For a little more detail, given $varepsilon>0$ with $c=r(a)+varepsilon < 1$, there exists $n_0$ such that for all $ngeq n_0$ we have $$ |r(a)-|a^*na^n|^1/2n|<varepsilon.$$ In particular,
$$
|a^*na^n|^1/2n<r(a)+varepsilon = c,
$$
which gives
$$
|a^*na^n|leq c^2n.
$$
answered Jul 28 at 19:58
Martin Argerami
115k1071164
edited Jul 29 at 4:40
answered Jul 28 at 19:58
Martin Argerami
115k1071164
answered Jul 28 at 19:58
Martin Argerami
115k1071164
answered Jul 28 at 19:58
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
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