Weak convergence of bounded sequence $(x_n)$ in Hilbert space where $langle{x_n,yrangle}rightarrow...
Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, the $x_n$ converges to $x$ weakly.
My Attempt
Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).
begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}
Is this correct please?
functional-analysis hilbert-spaces weak-convergence
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
add a comment |
Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, the $x_n$ converges to $x$ weakly.
My Attempt
Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).
begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}
Is this correct please?
functional-analysis hilbert-spaces weak-convergence
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
add a comment |
Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, the $x_n$ converges to $x$ weakly.
My Attempt
Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).
begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}
Is this correct please?
functional-analysis hilbert-spaces weak-convergence
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, the $x_n$ converges to $x$ weakly.
My Attempt
Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).
begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}
Is this correct please?
functional-analysis hilbert-spaces weak-convergence
functional-analysis hilbert-spaces weak-convergence
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
edited 2 days ago
Muhammad Mubarak
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
asked 2 days ago
Muhammad MubarakMuhammad Mubarak
598
598
add a comment |
add a comment |
1 Answer
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Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
$$
K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
$$
|langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
$$ where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
$$
limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
$$ Let $jtoinfty$ to conclude
$$
limsup_n |langle x-x_n,yrangle|=0,
$$ that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.
Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.
answered 2 days ago
SongSong
6,040318
add a comment |
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1 Answer
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1 Answer
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Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
$$
K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
$$
|langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
$$ where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
$$
limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
$$ Let $jtoinfty$ to conclude
$$
limsup_n |langle x-x_n,yrangle|=0,
$$ that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.
Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.
answered 2 days ago
SongSong
6,040318
add a comment |
Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
$$
K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
$$
|langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
$$ where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
$$
limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
$$ Let $jtoinfty$ to conclude
$$
limsup_n |langle x-x_n,yrangle|=0,
$$ that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.
Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.
answered 2 days ago
SongSong
6,040318
add a comment |
Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
$$
K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
$$
|langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
$$ where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
$$
limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
$$ Let $jtoinfty$ to conclude
$$
limsup_n |langle x-x_n,yrangle|=0,
$$ that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.
Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.
answered 2 days ago
SongSong
6,040318
Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
$$
K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
$$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
$$
|langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
$$ where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
$$
limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
$$ Let $jtoinfty$ to conclude
$$
limsup_n |langle x-x_n,yrangle|=0,
$$ that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.
Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.
answered 2 days ago
SongSong
6,040318
edited 2 days ago
answered 2 days ago
SongSong
6,040318
answered 2 days ago
SongSong
6,040318
answered 2 days ago
SongSong
6,040318
6,040318
add a comment |
add a comment |
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