Bounded operator on weakly convergent sequence maps to weakly convergent sequence
$begingroup$
Suppose that $H, K$ are Hilbert Spaces. We have a sequence $v_nin H$ that converges weakly to some vector $vin H$, that is:
$$langle v_n - v, wrangle to 0 quad text{as} quad ntoinfty$$
Also, we know that the operator $T:H to K$ is bounded.
I need to show that the sequence $Tv_nin K$ converges weakly to $Tvin K$.
My Proof
I tried proving it like this:
Let $epsilon>0$. Then since $v_n$ converges weakly to $v$ we have that there exists $n_0in mathbb{N}$ such that:
$$n>n_0 Longrightarrow |langle v_n-v, wrangle | < epsilon $$
Now, let $zin K$:
begin{align}
|langle Tv_n-Tv, zrangle| &= |langle T(v_n-v), zrangle| && text{as $T$ is linear}\
&= |langle v_n-v, T^*zrangle| && text{as every bounded operator has a unique adjoint $T^*$}\
&< epsilon && text{as $v_n$ converges weakly to $v$}
end{align}
Is this okay?
real-analysis linear-algebra functional-analysis hilbert-spaces
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
$endgroup$
add a comment |
$begingroup$
Suppose that $H, K$ are Hilbert Spaces. We have a sequence $v_nin H$ that converges weakly to some vector $vin H$, that is:
$$langle v_n - v, wrangle to 0 quad text{as} quad ntoinfty$$
Also, we know that the operator $T:H to K$ is bounded.
I need to show that the sequence $Tv_nin K$ converges weakly to $Tvin K$.
My Proof
I tried proving it like this:
Let $epsilon>0$. Then since $v_n$ converges weakly to $v$ we have that there exists $n_0in mathbb{N}$ such that:
$$n>n_0 Longrightarrow |langle v_n-v, wrangle | < epsilon $$
Now, let $zin K$:
begin{align}
|langle Tv_n-Tv, zrangle| &= |langle T(v_n-v), zrangle| && text{as $T$ is linear}\
&= |langle v_n-v, T^*zrangle| && text{as every bounded operator has a unique adjoint $T^*$}\
&< epsilon && text{as $v_n$ converges weakly to $v$}
end{align}
Is this okay?
real-analysis linear-algebra functional-analysis hilbert-spaces
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
$endgroup$
$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
1
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32
add a comment |
$begingroup$
Suppose that $H, K$ are Hilbert Spaces. We have a sequence $v_nin H$ that converges weakly to some vector $vin H$, that is:
$$langle v_n - v, wrangle to 0 quad text{as} quad ntoinfty$$
Also, we know that the operator $T:H to K$ is bounded.
I need to show that the sequence $Tv_nin K$ converges weakly to $Tvin K$.
My Proof
I tried proving it like this:
Let $epsilon>0$. Then since $v_n$ converges weakly to $v$ we have that there exists $n_0in mathbb{N}$ such that:
$$n>n_0 Longrightarrow |langle v_n-v, wrangle | < epsilon $$
Now, let $zin K$:
begin{align}
|langle Tv_n-Tv, zrangle| &= |langle T(v_n-v), zrangle| && text{as $T$ is linear}\
&= |langle v_n-v, T^*zrangle| && text{as every bounded operator has a unique adjoint $T^*$}\
&< epsilon && text{as $v_n$ converges weakly to $v$}
end{align}
Is this okay?
real-analysis linear-algebra functional-analysis hilbert-spaces
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
$endgroup$
Suppose that $H, K$ are Hilbert Spaces. We have a sequence $v_nin H$ that converges weakly to some vector $vin H$, that is:
$$langle v_n - v, wrangle to 0 quad text{as} quad ntoinfty$$
Also, we know that the operator $T:H to K$ is bounded.
I need to show that the sequence $Tv_nin K$ converges weakly to $Tvin K$.
My Proof
I tried proving it like this:
Let $epsilon>0$. Then since $v_n$ converges weakly to $v$ we have that there exists $n_0in mathbb{N}$ such that:
$$n>n_0 Longrightarrow |langle v_n-v, wrangle | < epsilon $$
Now, let $zin K$:
begin{align}
|langle Tv_n-Tv, zrangle| &= |langle T(v_n-v), zrangle| && text{as $T$ is linear}\
&= |langle v_n-v, T^*zrangle| && text{as every bounded operator has a unique adjoint $T^*$}\
&< epsilon && text{as $v_n$ converges weakly to $v$}
end{align}
Is this okay?
real-analysis linear-algebra functional-analysis hilbert-spaces
real-analysis linear-algebra functional-analysis hilbert-spaces
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
asked Dec 3 '18 at 23:10
Euler_SalterEuler_Salter
2,0571335
2,0571335
$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
1
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32
add a comment |
$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
1
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32
$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
1
1
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32
add a comment |
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$begingroup$
Actually, I think it is all wrong cause I am assuming it is linear
$endgroup$
– Euler_Salter
Dec 3 '18 at 23:12
1
$begingroup$
The term 'bounded operator' usually refers to a linear and continuous map. So your argument is fine.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:32