Mixing
Apparent (?) contraddiction with bounded operators
Clash Royale CLAN TAG#URR8PPP
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In mathematics, we can define
$l_2(mathbbN)= $$ $$x_i$$,hspace2mm $with$ hspace2mm x_i in mathbbN .
The following theorem can be simply proved:
Consider an operator $hatT in mathbbB$. If $hatT geq 0$,b then $hatT=hatT^*$.
I have to prove the following result:
Let $hatTinmathbbL(l_2(mathbbN)) | forall $$x_n$$in l_2(mathbbN)$, $nin mathbbN$, $hatT(x_n) = x_n+1/n$.
Prove that $hatT$ is bounded and calculate $hatT^*$.
Note that I used the notation: $mathbbL()$ for the linearity, $mathbbB$ for the bounded operators'set and $ hatT^*$ for the adjoint.
Well, the first proof is trivial and if I calculate the adjoint I discover that $hatT neq hatT^*$.
But, it is simple to show that $hatT geq 0$ and using the theorem it should result $hatT = hatT^*.$ Why I obtained 2 different results? Could anyone show me where I did wrong?
mathematical-physics
migrated from physics.stackexchange.com Jul 24 at 12:57
This question came from our site for active researchers, academics and students of physics.
add a comment |Â
up vote
-1
down vote
favorite
In mathematics, we can define
$l_2(mathbbN)= $$ $$x_i$$,hspace2mm $with$ hspace2mm x_i in mathbbN .
The following theorem can be simply proved:
Consider an operator $hatT in mathbbB$. If $hatT geq 0$,b then $hatT=hatT^*$.
I have to prove the following result:
Let $hatTinmathbbL(l_2(mathbbN)) | forall $$x_n$$in l_2(mathbbN)$, $nin mathbbN$, $hatT(x_n) = x_n+1/n$.
Prove that $hatT$ is bounded and calculate $hatT^*$.
Note that I used the notation: $mathbbL()$ for the linearity, $mathbbB$ for the bounded operators'set and $ hatT^*$ for the adjoint.
Well, the first proof is trivial and if I calculate the adjoint I discover that $hatT neq hatT^*$.
But, it is simple to show that $hatT geq 0$ and using the theorem it should result $hatT = hatT^*.$ Why I obtained 2 different results? Could anyone show me where I did wrong?
mathematical-physics
migrated from physics.stackexchange.com Jul 24 at 12:57
This question came from our site for active researchers, academics and students of physics.
$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
2
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In mathematics, we can define
$l_2(mathbbN)= $$ $$x_i$$,hspace2mm $with$ hspace2mm x_i in mathbbN .
The following theorem can be simply proved:
Consider an operator $hatT in mathbbB$. If $hatT geq 0$,b then $hatT=hatT^*$.
I have to prove the following result:
Let $hatTinmathbbL(l_2(mathbbN)) | forall $$x_n$$in l_2(mathbbN)$, $nin mathbbN$, $hatT(x_n) = x_n+1/n$.
Prove that $hatT$ is bounded and calculate $hatT^*$.
Note that I used the notation: $mathbbL()$ for the linearity, $mathbbB$ for the bounded operators'set and $ hatT^*$ for the adjoint.
Well, the first proof is trivial and if I calculate the adjoint I discover that $hatT neq hatT^*$.
But, it is simple to show that $hatT geq 0$ and using the theorem it should result $hatT = hatT^*.$ Why I obtained 2 different results? Could anyone show me where I did wrong?
mathematical-physics
In mathematics, we can define
$l_2(mathbbN)= $$ $$x_i$$,hspace2mm $with$ hspace2mm x_i in mathbbN .
The following theorem can be simply proved:
Consider an operator $hatT in mathbbB$. If $hatT geq 0$,b then $hatT=hatT^*$.
I have to prove the following result:
Let $hatTinmathbbL(l_2(mathbbN)) | forall $$x_n$$in l_2(mathbbN)$, $nin mathbbN$, $hatT(x_n) = x_n+1/n$.
Prove that $hatT$ is bounded and calculate $hatT^*$.
Note that I used the notation: $mathbbL()$ for the linearity, $mathbbB$ for the bounded operators'set and $ hatT^*$ for the adjoint.
Well, the first proof is trivial and if I calculate the adjoint I discover that $hatT neq hatT^*$.
But, it is simple to show that $hatT geq 0$ and using the theorem it should result $hatT = hatT^*.$ Why I obtained 2 different results? Could anyone show me where I did wrong?
mathematical-physics
asked Jul 23 at 23:47
SimonTat
migrated from physics.stackexchange.com Jul 24 at 12:57
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jul 24 at 12:57
This question came from our site for active researchers, academics and students of physics.
$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
2
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07
add a comment |Â
$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
2
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07
$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
2
2
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07
add a comment |Â
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$hatT$ is not symmetric so you can't have $hatT geq 0$.
– Keith McClary
Jul 24 at 1:55
$hatT in mathbbL(H, H) geq 0$ if $forall psi in H, (psi, hatTpsi) geq 0$
– SimonTat
Jul 24 at 5:55
2
If $x=(0,-1,1,0,0,...)$, then $hatT(x)=(frac11,0,0,...)$. If follows that $left(x,hatTxright)=-1<0$. From the statement I didn't understand if $hatT$ is a left or right weighted shift. If it shifts to the right, then take $x=(-1,1,0,...)$ and do the same computation.
– user578878
Jul 24 at 13:07