Show that multiplication operator in $L^infty$ is self adjoint
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I am trying to show that the following operator is self adjoint.
$$T:L^infty[0,1]rightarrow L^infty[0,1]$$ such that $$(Tx)(t)=tx(t)$$
I am stuck mainly because I can't seem to find how the inner product is defined in $L^infty$. I know that in $L^2$ we have that $$langle Tx,yrangle = int_0^1 (Txbary)^2$$
Is there any equivalence to this for $L^infty$?
functional-analysis
asked Aug 10 at 23:09
MathIsHard
1,149415
 |Â
show 2 more comments
up vote
0
down vote
favorite
I am trying to show that the following operator is self adjoint.
$$T:L^infty[0,1]rightarrow L^infty[0,1]$$ such that $$(Tx)(t)=tx(t)$$
I am stuck mainly because I can't seem to find how the inner product is defined in $L^infty$. I know that in $L^2$ we have that $$langle Tx,yrangle = int_0^1 (Txbary)^2$$
Is there any equivalence to this for $L^infty$?
functional-analysis
asked Aug 10 at 23:09
MathIsHard
1,149415
2
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
1
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
1
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
1
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to show that the following operator is self adjoint.
$$T:L^infty[0,1]rightarrow L^infty[0,1]$$ such that $$(Tx)(t)=tx(t)$$
I am stuck mainly because I can't seem to find how the inner product is defined in $L^infty$. I know that in $L^2$ we have that $$langle Tx,yrangle = int_0^1 (Txbary)^2$$
Is there any equivalence to this for $L^infty$?
functional-analysis
asked Aug 10 at 23:09
MathIsHard
1,149415
I am trying to show that the following operator is self adjoint.
$$T:L^infty[0,1]rightarrow L^infty[0,1]$$ such that $$(Tx)(t)=tx(t)$$
I am stuck mainly because I can't seem to find how the inner product is defined in $L^infty$. I know that in $L^2$ we have that $$langle Tx,yrangle = int_0^1 (Txbary)^2$$
Is there any equivalence to this for $L^infty$?
functional-analysis
asked Aug 10 at 23:09
MathIsHard
1,149415
asked Aug 10 at 23:09
MathIsHard
1,149415
asked Aug 10 at 23:09
MathIsHard
1,149415
asked Aug 10 at 23:09
MathIsHard
1,149415
1,149415
2
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
1
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
1
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
1
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09
 |Â
show 2 more comments
2
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
1
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
1
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
1
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09
2
2
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
1
1
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
1
1
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
1
1
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09
 |Â
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2
That problem has no solution, since the norm $|cdot|_infty$ is not induced by an inner product.
– José Carlos Santos
Aug 10 at 23:14
Oh I see. Ok, thank you for the help on that. I am really trying to find that the spectrum = the point spectrum = $[0,1]$. I showed that the point spectrum is $[0,1]$ and I showed that all $|lambda|leq ||T||=1$ but I am stuck on how to show that there are no other spectral values. I was thinking that if I could show it was self adjoint then the spectral values are all real. But maybe it is enough since the space is $L^infty[0,1]$...
– MathIsHard
Aug 10 at 23:23
1
Just try to explicitly write down an inverse of $T-lambda$ if $lambdanotin[0,1]$.
– Eric Wofsey
Aug 10 at 23:25
1
@JoséCarlosSantos The norm doesn't need to come from an inner product to be able to talk about adjoints, self-adjoint, etc. The real problem is that $L^infty$ is not its own dual. If you take a self-dual Banach space, you can perfectly consider operators in them that are self-adjoint.
– user583012
Aug 10 at 23:31
1
It makes no sense to say that this $T$ is self-adjoint. Because if $T:Xto YY$ then $T^*:Y^*to X^*$, so here $T=T*$ is simply impossible; $T$ and $T^*$ are defined on different spaces. An operator from $L^2$ to itself can be self-adjoint, because $(L^2)^*=L^2$. (Regarding "I found it here:". No, the $T$ in that other post is a different operator. Because it's defined on $L^2$, while the current one is defined on $L^infty$.)
– David C. Ullrich
Aug 11 at 0:09