Operator Norm Question
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Suppose I am interested in operators $T:Xto Y$, with $X$ and $Y$ both separable Hilbert spaces. The operator norm of such $T$ can then be taken as
$$
|T| = sup_{|x|_Xleq 1}|Tx|_Y.
$$
Since the spaces are separable, there exists a sequence ${x_n}$ and ${y_m}$ dense in the unit balls of $X$ and $Y$. It would appear then, that we can write
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
In the above, I am representing the $Y$ norm as
$$
|y|_Y = sup_{|tilde y|leq 1}|(y,tilde y)_Y|
$$
Is there anything wrong with this intuition?
functional-analysis operator-theory
asked yesterday
user2379888
1746
add a comment |
up vote
2
down vote
favorite
Suppose I am interested in operators $T:Xto Y$, with $X$ and $Y$ both separable Hilbert spaces. The operator norm of such $T$ can then be taken as
$$
|T| = sup_{|x|_Xleq 1}|Tx|_Y.
$$
Since the spaces are separable, there exists a sequence ${x_n}$ and ${y_m}$ dense in the unit balls of $X$ and $Y$. It would appear then, that we can write
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
In the above, I am representing the $Y$ norm as
$$
|y|_Y = sup_{|tilde y|leq 1}|(y,tilde y)_Y|
$$
Is there anything wrong with this intuition?
functional-analysis operator-theory
asked yesterday
user2379888
1746
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I am interested in operators $T:Xto Y$, with $X$ and $Y$ both separable Hilbert spaces. The operator norm of such $T$ can then be taken as
$$
|T| = sup_{|x|_Xleq 1}|Tx|_Y.
$$
Since the spaces are separable, there exists a sequence ${x_n}$ and ${y_m}$ dense in the unit balls of $X$ and $Y$. It would appear then, that we can write
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
In the above, I am representing the $Y$ norm as
$$
|y|_Y = sup_{|tilde y|leq 1}|(y,tilde y)_Y|
$$
Is there anything wrong with this intuition?
functional-analysis operator-theory
asked yesterday
user2379888
1746
Suppose I am interested in operators $T:Xto Y$, with $X$ and $Y$ both separable Hilbert spaces. The operator norm of such $T$ can then be taken as
$$
|T| = sup_{|x|_Xleq 1}|Tx|_Y.
$$
Since the spaces are separable, there exists a sequence ${x_n}$ and ${y_m}$ dense in the unit balls of $X$ and $Y$. It would appear then, that we can write
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
In the above, I am representing the $Y$ norm as
$$
|y|_Y = sup_{|tilde y|leq 1}|(y,tilde y)_Y|
$$
Is there anything wrong with this intuition?
functional-analysis operator-theory
functional-analysis operator-theory
asked yesterday
user2379888
1746
asked yesterday
user2379888
1746
asked yesterday
user2379888
1746
asked yesterday
user2379888
1746
asked yesterday
user2379888
1746
1746
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your equation
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
is indeed correct.
It works because the relevant norms are continuous and because of the density of the sequences in the unit balls.
answered yesterday
supinf
5,565926
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your equation
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
is indeed correct.
It works because the relevant norms are continuous and because of the density of the sequences in the unit balls.
answered yesterday
supinf
5,565926
add a comment |
up vote
1
down vote
accepted
Your equation
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
is indeed correct.
It works because the relevant norms are continuous and because of the density of the sequences in the unit balls.
answered yesterday
supinf
5,565926
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your equation
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
is indeed correct.
It works because the relevant norms are continuous and because of the density of the sequences in the unit balls.
answered yesterday
supinf
5,565926
Your equation
$$
|T|=sup_{n}sup_m |(Tx_n,y_m)_Y| = sup_{m}sup_n |(Tx_n,y_m)_Y|.
$$
is indeed correct.
It works because the relevant norms are continuous and because of the density of the sequences in the unit balls.
answered yesterday
supinf
5,565926
answered yesterday
supinf
5,565926
answered yesterday
supinf
5,565926
answered yesterday
supinf
5,565926
5,565926
add a comment |
add a comment |
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