Mixing
norm in separable Hilbert space as countable supremum
$begingroup$
If $H$ is an Hilbert space with inner product denoted by $(cdot , cdot )$, then
$$||u||=sup {(u,v)::||v|| leq 1 }$$
Question: In the case that $H$ is separable, is there any countable family ${ w_n} subset H$ such that $||u||=sup_{n in mathbb N} , (u,w_n)$ ?
I know that in this case $H$ admists an orthonormal basis and the Parseval identity holds, but I don't know how to use it
functional-analysis hilbert-spaces
asked Jan 25 at 13:27
LouisLouis
1247
$endgroup$
add a comment |
$begingroup$
If $H$ is an Hilbert space with inner product denoted by $(cdot , cdot )$, then
$$||u||=sup {(u,v)::||v|| leq 1 }$$
Question: In the case that $H$ is separable, is there any countable family ${ w_n} subset H$ such that $||u||=sup_{n in mathbb N} , (u,w_n)$ ?
I know that in this case $H$ admists an orthonormal basis and the Parseval identity holds, but I don't know how to use it
functional-analysis hilbert-spaces
asked Jan 25 at 13:27
LouisLouis
1247
$endgroup$
1
$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
2
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45
add a comment |
$begingroup$
If $H$ is an Hilbert space with inner product denoted by $(cdot , cdot )$, then
$$||u||=sup {(u,v)::||v|| leq 1 }$$
Question: In the case that $H$ is separable, is there any countable family ${ w_n} subset H$ such that $||u||=sup_{n in mathbb N} , (u,w_n)$ ?
I know that in this case $H$ admists an orthonormal basis and the Parseval identity holds, but I don't know how to use it
functional-analysis hilbert-spaces
asked Jan 25 at 13:27
LouisLouis
1247
$endgroup$
If $H$ is an Hilbert space with inner product denoted by $(cdot , cdot )$, then
$$||u||=sup {(u,v)::||v|| leq 1 }$$
Question: In the case that $H$ is separable, is there any countable family ${ w_n} subset H$ such that $||u||=sup_{n in mathbb N} , (u,w_n)$ ?
I know that in this case $H$ admists an orthonormal basis and the Parseval identity holds, but I don't know how to use it
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Jan 25 at 13:27
LouisLouis
1247
asked Jan 25 at 13:27
LouisLouis
1247
asked Jan 25 at 13:27
LouisLouis
1247
asked Jan 25 at 13:27
LouisLouis
1247
asked Jan 25 at 13:27
LouisLouis
1247
1247
1
$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
2
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45
add a comment |
1
$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
2
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45
1
1
$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
2
2
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis ${e_i}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1={w_k:k in mathbb N}$ the set of elements of $E$ with norm at most $1$.
Fix $u in H$. I want to prove that
$$sup{(u,v):,v in H, ||v||leq 1 }=sup_{k in mathbb N},(u,w_k)$$
and since one inequality is trivial I only need to prove this "$le$"
For any $v in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n to v$ in $H$. If $||v|| leq 1$, then $v_n in E_1$ for $n >>1$. Given $epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||leq epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)geq (u,v)-epsilon$$
thus $sup_{k in mathbb N},(u,w_k) geq (u,v)- epsilon$. Taking the supremum over $v in H$ with $||v||leq 1$ and letting $ epsilon to 0^+$, the claim follows.
answered Jan 26 at 11:08
LouisLouis
1247
$endgroup$
add a comment |
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$begingroup$
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis ${e_i}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1={w_k:k in mathbb N}$ the set of elements of $E$ with norm at most $1$.
Fix $u in H$. I want to prove that
$$sup{(u,v):,v in H, ||v||leq 1 }=sup_{k in mathbb N},(u,w_k)$$
and since one inequality is trivial I only need to prove this "$le$"
For any $v in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n to v$ in $H$. If $||v|| leq 1$, then $v_n in E_1$ for $n >>1$. Given $epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||leq epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)geq (u,v)-epsilon$$
thus $sup_{k in mathbb N},(u,w_k) geq (u,v)- epsilon$. Taking the supremum over $v in H$ with $||v||leq 1$ and letting $ epsilon to 0^+$, the claim follows.
answered Jan 26 at 11:08
LouisLouis
1247
$endgroup$
add a comment |
$begingroup$
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis ${e_i}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1={w_k:k in mathbb N}$ the set of elements of $E$ with norm at most $1$.
Fix $u in H$. I want to prove that
$$sup{(u,v):,v in H, ||v||leq 1 }=sup_{k in mathbb N},(u,w_k)$$
and since one inequality is trivial I only need to prove this "$le$"
For any $v in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n to v$ in $H$. If $||v|| leq 1$, then $v_n in E_1$ for $n >>1$. Given $epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||leq epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)geq (u,v)-epsilon$$
thus $sup_{k in mathbb N},(u,w_k) geq (u,v)- epsilon$. Taking the supremum over $v in H$ with $||v||leq 1$ and letting $ epsilon to 0^+$, the claim follows.
answered Jan 26 at 11:08
LouisLouis
1247
$endgroup$
add a comment |
$begingroup$
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis ${e_i}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1={w_k:k in mathbb N}$ the set of elements of $E$ with norm at most $1$.
Fix $u in H$. I want to prove that
$$sup{(u,v):,v in H, ||v||leq 1 }=sup_{k in mathbb N},(u,w_k)$$
and since one inequality is trivial I only need to prove this "$le$"
For any $v in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n to v$ in $H$. If $||v|| leq 1$, then $v_n in E_1$ for $n >>1$. Given $epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||leq epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)geq (u,v)-epsilon$$
thus $sup_{k in mathbb N},(u,w_k) geq (u,v)- epsilon$. Taking the supremum over $v in H$ with $||v||leq 1$ and letting $ epsilon to 0^+$, the claim follows.
answered Jan 26 at 11:08
LouisLouis
1247
$endgroup$
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis ${e_i}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1={w_k:k in mathbb N}$ the set of elements of $E$ with norm at most $1$.
Fix $u in H$. I want to prove that
$$sup{(u,v):,v in H, ||v||leq 1 }=sup_{k in mathbb N},(u,w_k)$$
and since one inequality is trivial I only need to prove this "$le$"
For any $v in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n to v$ in $H$. If $||v|| leq 1$, then $v_n in E_1$ for $n >>1$. Given $epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||leq epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)geq (u,v)-epsilon$$
thus $sup_{k in mathbb N},(u,w_k) geq (u,v)- epsilon$. Taking the supremum over $v in H$ with $||v||leq 1$ and letting $ epsilon to 0^+$, the claim follows.
answered Jan 26 at 11:08
LouisLouis
1247
answered Jan 26 at 11:08
LouisLouis
1247
answered Jan 26 at 11:08
LouisLouis
1247
answered Jan 26 at 11:08
LouisLouis
1247
1247
add a comment |
add a comment |
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$begingroup$
How about all possible finite rational linear combinations of some countable orthonormal set? Would that work? I mean, consider a countable orthonormal basis $u_n$ on $H$, and consider the set ${x : x = sum x_nu_n, |x_n|^2 leq 1, x_n in mathbb Q forall n} = S$. Try to see if this works i.e. check if $||u|| = sup_{s in S} langle u,srangle$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 25 at 13:35
2
$begingroup$
Just take a countable dense subset of the unit ball?
$endgroup$
– trii
Jan 25 at 14:45