Mixing
span density of shift operator
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Let $mathcalH$ be a Hilbert space with complete orthonormal basis $ nin mathbbN$. Let $R$ be the shift operator on $mathcalH$ defined by $R(e_n)=e_n+1$, and extended by linearity and continuity. Show that there exists no $xin mathcalH$, such that $textspanR^2n(x)$ is dense in $mathcalH$.
Assuming the contrary, we get a $xin mathcalH$ such that $R^2n(x)$ is spanned over $mathcalH$ for some $ngeq 0$, i.e. $R^2n(x)=c_1e_1+c_2e_2+dots +c_ne_n$ for all $c_iin mathbbR$. Taking $R$ operated on both sides, $R^2n+1(x)=c_1R(e_1)+c_2R(e_2)+dots +c_nR(e_n)=c_1e_2+c_2e_3+dots +c_ne_n+1$, since $R$ is extended by linearity. Proceeding as above, after $n$-times we get, $R^3n(x)=c_1e_n+1+c_2e_n+2+dots +c_ne_2n$. But from here, I'm unable to figure out what to do, specially how to use the fact that $R$ is extended by continuity, and all $e_i$'s are orthonormal. How can I bring the contradiction by the above process?
functional-analysis
asked Jul 27 at 17:42
am_11235...
1797
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up vote
2
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Let $mathcalH$ be a Hilbert space with complete orthonormal basis $ nin mathbbN$. Let $R$ be the shift operator on $mathcalH$ defined by $R(e_n)=e_n+1$, and extended by linearity and continuity. Show that there exists no $xin mathcalH$, such that $textspanR^2n(x)$ is dense in $mathcalH$.
Assuming the contrary, we get a $xin mathcalH$ such that $R^2n(x)$ is spanned over $mathcalH$ for some $ngeq 0$, i.e. $R^2n(x)=c_1e_1+c_2e_2+dots +c_ne_n$ for all $c_iin mathbbR$. Taking $R$ operated on both sides, $R^2n+1(x)=c_1R(e_1)+c_2R(e_2)+dots +c_nR(e_n)=c_1e_2+c_2e_3+dots +c_ne_n+1$, since $R$ is extended by linearity. Proceeding as above, after $n$-times we get, $R^3n(x)=c_1e_n+1+c_2e_n+2+dots +c_ne_2n$. But from here, I'm unable to figure out what to do, specially how to use the fact that $R$ is extended by continuity, and all $e_i$'s are orthonormal. How can I bring the contradiction by the above process?
functional-analysis
asked Jul 27 at 17:42
am_11235...
1797
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $mathcalH$ be a Hilbert space with complete orthonormal basis $ nin mathbbN$. Let $R$ be the shift operator on $mathcalH$ defined by $R(e_n)=e_n+1$, and extended by linearity and continuity. Show that there exists no $xin mathcalH$, such that $textspanR^2n(x)$ is dense in $mathcalH$.
Assuming the contrary, we get a $xin mathcalH$ such that $R^2n(x)$ is spanned over $mathcalH$ for some $ngeq 0$, i.e. $R^2n(x)=c_1e_1+c_2e_2+dots +c_ne_n$ for all $c_iin mathbbR$. Taking $R$ operated on both sides, $R^2n+1(x)=c_1R(e_1)+c_2R(e_2)+dots +c_nR(e_n)=c_1e_2+c_2e_3+dots +c_ne_n+1$, since $R$ is extended by linearity. Proceeding as above, after $n$-times we get, $R^3n(x)=c_1e_n+1+c_2e_n+2+dots +c_ne_2n$. But from here, I'm unable to figure out what to do, specially how to use the fact that $R$ is extended by continuity, and all $e_i$'s are orthonormal. How can I bring the contradiction by the above process?
functional-analysis
asked Jul 27 at 17:42
am_11235...
1797
Let $mathcalH$ be a Hilbert space with complete orthonormal basis $ nin mathbbN$. Let $R$ be the shift operator on $mathcalH$ defined by $R(e_n)=e_n+1$, and extended by linearity and continuity. Show that there exists no $xin mathcalH$, such that $textspanR^2n(x)$ is dense in $mathcalH$.
Assuming the contrary, we get a $xin mathcalH$ such that $R^2n(x)$ is spanned over $mathcalH$ for some $ngeq 0$, i.e. $R^2n(x)=c_1e_1+c_2e_2+dots +c_ne_n$ for all $c_iin mathbbR$. Taking $R$ operated on both sides, $R^2n+1(x)=c_1R(e_1)+c_2R(e_2)+dots +c_nR(e_n)=c_1e_2+c_2e_3+dots +c_ne_n+1$, since $R$ is extended by linearity. Proceeding as above, after $n$-times we get, $R^3n(x)=c_1e_n+1+c_2e_n+2+dots +c_ne_2n$. But from here, I'm unable to figure out what to do, specially how to use the fact that $R$ is extended by continuity, and all $e_i$'s are orthonormal. How can I bring the contradiction by the above process?
functional-analysis
asked Jul 27 at 17:42
am_11235...
1797
asked Jul 27 at 17:42
am_11235...
1797
asked Jul 27 at 17:42
am_11235...
1797
asked Jul 27 at 17:42
am_11235...
1797
1797
add a comment |Â
add a comment |Â
2 Answers
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I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here.
If you write $x=sum_j x_je_j$ and you consider a linear combination $z=sum_n=0^k c_nR^2nx$, you have
$$
z=sum_n=0^k c_nsum_j x_j e_j+2n=sum_jsum_n=0^kc_nx_je_j+2n.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with
$$
z=c_0x_1e_1+c_0x_2e_2+cdots
$$
For any $y=y_1e_1+y_2e_2+cdots$, we have $$|y-z|^2geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then
$$
|y-z|^2geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=minx_2$. If $r>0$, we have
$$
|y-z|^2geq r,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $|e_1-z|^2geq1$; and similarly, if $x_2=0$, then $e_2-z|^2geq1$.
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,ldots$
answered Jul 27 at 18:14
Martin Argerami
115k1071164
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Let $x in H, x ne 0$ and let $n in mathbbN$ be the smallest number such that $langle x, e_2n-1rangle ne 0$.
Consider $y = langle x, e_2nrangle e_2n-1- langle x, e_2n-1rangle e_2n$. Clearly $y ne 0$.
For $k = 0$ we have $langle y, xrangle = 0$ and for $k ge 1$ we have $R^2n(x) in operatornamespane_k_k ge 2n+1$ so again $langle y, xrangle = 0$.
Therefore $y perp R^2n(x) : n ge 0$ so it cannot be dense in $H$.
answered Jul 27 at 22:05
mechanodroid
22.2k52041
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here.
If you write $x=sum_j x_je_j$ and you consider a linear combination $z=sum_n=0^k c_nR^2nx$, you have
$$
z=sum_n=0^k c_nsum_j x_j e_j+2n=sum_jsum_n=0^kc_nx_je_j+2n.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with
$$
z=c_0x_1e_1+c_0x_2e_2+cdots
$$
For any $y=y_1e_1+y_2e_2+cdots$, we have $$|y-z|^2geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then
$$
|y-z|^2geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=minx_2$. If $r>0$, we have
$$
|y-z|^2geq r,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $|e_1-z|^2geq1$; and similarly, if $x_2=0$, then $e_2-z|^2geq1$.
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,ldots$
answered Jul 27 at 18:14
Martin Argerami
115k1071164
add a comment |Â
up vote
0
down vote
I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here.
If you write $x=sum_j x_je_j$ and you consider a linear combination $z=sum_n=0^k c_nR^2nx$, you have
$$
z=sum_n=0^k c_nsum_j x_j e_j+2n=sum_jsum_n=0^kc_nx_je_j+2n.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with
$$
z=c_0x_1e_1+c_0x_2e_2+cdots
$$
For any $y=y_1e_1+y_2e_2+cdots$, we have $$|y-z|^2geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then
$$
|y-z|^2geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=minx_2$. If $r>0$, we have
$$
|y-z|^2geq r,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $|e_1-z|^2geq1$; and similarly, if $x_2=0$, then $e_2-z|^2geq1$.
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,ldots$
answered Jul 27 at 18:14
Martin Argerami
115k1071164
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here.
If you write $x=sum_j x_je_j$ and you consider a linear combination $z=sum_n=0^k c_nR^2nx$, you have
$$
z=sum_n=0^k c_nsum_j x_j e_j+2n=sum_jsum_n=0^kc_nx_je_j+2n.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with
$$
z=c_0x_1e_1+c_0x_2e_2+cdots
$$
For any $y=y_1e_1+y_2e_2+cdots$, we have $$|y-z|^2geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then
$$
|y-z|^2geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=minx_2$. If $r>0$, we have
$$
|y-z|^2geq r,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $|e_1-z|^2geq1$; and similarly, if $x_2=0$, then $e_2-z|^2geq1$.
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,ldots$
answered Jul 27 at 18:14
Martin Argerami
115k1071164
I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here.
If you write $x=sum_j x_je_j$ and you consider a linear combination $z=sum_n=0^k c_nR^2nx$, you have
$$
z=sum_n=0^k c_nsum_j x_j e_j+2n=sum_jsum_n=0^kc_nx_je_j+2n.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with
$$
z=c_0x_1e_1+c_0x_2e_2+cdots
$$
For any $y=y_1e_1+y_2e_2+cdots$, we have $$|y-z|^2geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then
$$
|y-z|^2geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=minx_2$. If $r>0$, we have
$$
|y-z|^2geq r,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $|e_1-z|^2geq1$; and similarly, if $x_2=0$, then $e_2-z|^2geq1$.
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,ldots$
answered Jul 27 at 18:14
Martin Argerami
115k1071164
answered Jul 27 at 18:14
Martin Argerami
115k1071164
answered Jul 27 at 18:14
Martin Argerami
115k1071164
answered Jul 27 at 18:14
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $x in H, x ne 0$ and let $n in mathbbN$ be the smallest number such that $langle x, e_2n-1rangle ne 0$.
Consider $y = langle x, e_2nrangle e_2n-1- langle x, e_2n-1rangle e_2n$. Clearly $y ne 0$.
For $k = 0$ we have $langle y, xrangle = 0$ and for $k ge 1$ we have $R^2n(x) in operatornamespane_k_k ge 2n+1$ so again $langle y, xrangle = 0$.
Therefore $y perp R^2n(x) : n ge 0$ so it cannot be dense in $H$.
answered Jul 27 at 22:05
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
Let $x in H, x ne 0$ and let $n in mathbbN$ be the smallest number such that $langle x, e_2n-1rangle ne 0$.
Consider $y = langle x, e_2nrangle e_2n-1- langle x, e_2n-1rangle e_2n$. Clearly $y ne 0$.
For $k = 0$ we have $langle y, xrangle = 0$ and for $k ge 1$ we have $R^2n(x) in operatornamespane_k_k ge 2n+1$ so again $langle y, xrangle = 0$.
Therefore $y perp R^2n(x) : n ge 0$ so it cannot be dense in $H$.
answered Jul 27 at 22:05
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $x in H, x ne 0$ and let $n in mathbbN$ be the smallest number such that $langle x, e_2n-1rangle ne 0$.
Consider $y = langle x, e_2nrangle e_2n-1- langle x, e_2n-1rangle e_2n$. Clearly $y ne 0$.
For $k = 0$ we have $langle y, xrangle = 0$ and for $k ge 1$ we have $R^2n(x) in operatornamespane_k_k ge 2n+1$ so again $langle y, xrangle = 0$.
Therefore $y perp R^2n(x) : n ge 0$ so it cannot be dense in $H$.
answered Jul 27 at 22:05
mechanodroid
22.2k52041
Let $x in H, x ne 0$ and let $n in mathbbN$ be the smallest number such that $langle x, e_2n-1rangle ne 0$.
Consider $y = langle x, e_2nrangle e_2n-1- langle x, e_2n-1rangle e_2n$. Clearly $y ne 0$.
For $k = 0$ we have $langle y, xrangle = 0$ and for $k ge 1$ we have $R^2n(x) in operatornamespane_k_k ge 2n+1$ so again $langle y, xrangle = 0$.
Therefore $y perp R^2n(x) : n ge 0$ so it cannot be dense in $H$.
answered Jul 27 at 22:05
mechanodroid
22.2k52041
answered Jul 27 at 22:05
mechanodroid
22.2k52041
answered Jul 27 at 22:05
mechanodroid
22.2k52041
answered Jul 27 at 22:05
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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