If $f neq 0$ is a continuous functional linear on Hilbert space, is it possible that $f(v_n) to 0$ and $|v_n|...
Let $H$ be a infinite dimensional Hilbert space over the complex plane $mathbb{C}$
Let $f: H to mathbb{C}$ be a continuous linear functional on $H$ such that $f neq 0$
Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence in $H$ such that $|v_n| to infty$
My question: Is it possible that $f(v_n) to 0$ ?
Thanks.
functional-analysis convergence hilbert-spaces
asked Nov 29 at 14:54
Matey Math
839414
add a comment |
Let $H$ be a infinite dimensional Hilbert space over the complex plane $mathbb{C}$
Let $f: H to mathbb{C}$ be a continuous linear functional on $H$ such that $f neq 0$
Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence in $H$ such that $|v_n| to infty$
My question: Is it possible that $f(v_n) to 0$ ?
Thanks.
functional-analysis convergence hilbert-spaces
asked Nov 29 at 14:54
Matey Math
839414
2
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11
add a comment |
Let $H$ be a infinite dimensional Hilbert space over the complex plane $mathbb{C}$
Let $f: H to mathbb{C}$ be a continuous linear functional on $H$ such that $f neq 0$
Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence in $H$ such that $|v_n| to infty$
My question: Is it possible that $f(v_n) to 0$ ?
Thanks.
functional-analysis convergence hilbert-spaces
asked Nov 29 at 14:54
Matey Math
839414
Let $H$ be a infinite dimensional Hilbert space over the complex plane $mathbb{C}$
Let $f: H to mathbb{C}$ be a continuous linear functional on $H$ such that $f neq 0$
Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence in $H$ such that $|v_n| to infty$
My question: Is it possible that $f(v_n) to 0$ ?
Thanks.
functional-analysis convergence hilbert-spaces
functional-analysis convergence hilbert-spaces
asked Nov 29 at 14:54
Matey Math
839414
asked Nov 29 at 14:54
Matey Math
839414
asked Nov 29 at 14:54
Matey Math
839414
asked Nov 29 at 14:54
Matey Math
839414
asked Nov 29 at 14:54
Matey Math
839414
839414
2
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11
add a comment |
2
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11
2
2
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11
add a comment |
1 Answer
1
active
oldest
votes
It is possible. Choose $xinker f$, $ynotinker f$ normalized with $xperp y$. Let $z_n= nx+frac1ny$. Then $|z_n|toinfty$ and $f(z_n)=frac1nf(y)to0$.
answered Nov 29 at 17:10
Aweygan
13.4k21441
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is possible. Choose $xinker f$, $ynotinker f$ normalized with $xperp y$. Let $z_n= nx+frac1ny$. Then $|z_n|toinfty$ and $f(z_n)=frac1nf(y)to0$.
answered Nov 29 at 17:10
Aweygan
13.4k21441
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
add a comment |
It is possible. Choose $xinker f$, $ynotinker f$ normalized with $xperp y$. Let $z_n= nx+frac1ny$. Then $|z_n|toinfty$ and $f(z_n)=frac1nf(y)to0$.
answered Nov 29 at 17:10
Aweygan
13.4k21441
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
add a comment |
It is possible. Choose $xinker f$, $ynotinker f$ normalized with $xperp y$. Let $z_n= nx+frac1ny$. Then $|z_n|toinfty$ and $f(z_n)=frac1nf(y)to0$.
answered Nov 29 at 17:10
Aweygan
13.4k21441
It is possible. Choose $xinker f$, $ynotinker f$ normalized with $xperp y$. Let $z_n= nx+frac1ny$. Then $|z_n|toinfty$ and $f(z_n)=frac1nf(y)to0$.
answered Nov 29 at 17:10
Aweygan
13.4k21441
answered Nov 29 at 17:10
Aweygan
13.4k21441
answered Nov 29 at 17:10
Aweygan
13.4k21441
answered Nov 29 at 17:10
Aweygan
13.4k21441
13.4k21441
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
add a comment |
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
Thanks @Aweygan for your answer
– Matey Math
Nov 29 at 17:20
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
You're welcome. Glad to help!
– Aweygan
Nov 29 at 17:21
add a comment |
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2
Yes, if all $v_n$ are in $ker f$
– Aweygan
Nov 29 at 14:58
@Aweygan thanks for your comment, if no one $v_n$ is in $ker f$ ?
– Matey Math
Nov 29 at 15:11