Weak cauchy operator sequence over hilbert spaces
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
add a comment |
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
add a comment |
$begingroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
$endgroup$
Supose that $X$ and $Y$ are two Hilbert spaces and $(T_n)$ a weakly Cauchy sequence. That means for all $x in X$ and $y in Y$, $(langle x, T_ny rangle )_{n in mathbb{N}}$ is a Cauchy sequence in $mathbb{C}$. Prove that exist a linear bounded operator $T in mathcal{L}(X,Y)$ such that $T_n$ converges weakly to $T$. That means for all $x in X$ and $y in Y$ $langle x, T_ny rangle xrightarrow{n to infty } langle x, Ty rangle$.
I know if $(T_n)$ is weakly Cauchy sequence then $||T_n||$ is bounded (by Banach-Steinhaus and $||T_n||=sup_{||x||=||y||=1}|langle x, T_nyrangle |$) and $(langle x, T_ny rangle )_{n in mathbb{N}} xrightarrow{n to infty} L_{x,y}$ for some $L_{x,y}in mathbb{C}$. But I don't know how construct a linear operator $T in mathcal{L}(X,Y) $ such that $L_{x,y}=langle x, Ty rangle$. How can I construct that if is possible?
functional-analysis hilbert-spaces weak-convergence
functional-analysis hilbert-spaces weak-convergence
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
asked Dec 13 '18 at 3:01
Pablo HerreraPablo Herrera
399113
399113
add a comment |
add a comment |
1 Answer
1
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Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
13.1k632
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add a comment |
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$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
13.1k632
$endgroup$
add a comment |
$begingroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
13.1k632
$endgroup$
Riesz representation can do it. Note that
$$
xmapsto L_{x,y}
$$ defines a bounded linear functional from $X$ into $mathbb{C}$ for each $yin Y$ since it holds that
$$
|L_{x,y}|leq left(sup_n lVert T_nrVertright)lVert xrVert lVert yrVertquadcdots(*).
$$ By Riesz representation theorem, this implies that for each $yin Y$, there exists unique $xi_yin X$ such that
$$
L_{x,y} = langle x,xi_y rangle,
$$ and it holds that
$$
|xi_y| = lVert L_{cdot,y} rVertleq sup_n lVert T_nrVert cdot lVert yrVert,
$$ by $(*)$. Finally, note that $ymapsto xi_y$ is linear in $yin Y$. Therefore, by defining a bounded linear operator $T$ as
$$
T : yin Ymapsto xi_y in X,
$$ we have $L_{x,y} = lim_{ntoinfty} langle x,T_nyrangle = langle x,Tyrangle$ and $T$ is a weak limit of ${T_n}$.
answered Dec 13 '18 at 3:32
SongSong
13.1k632
answered Dec 13 '18 at 3:32
SongSong
13.1k632
answered Dec 13 '18 at 3:32
SongSong
13.1k632
answered Dec 13 '18 at 3:32
SongSong
13.1k632
13.1k632
add a comment |
add a comment |
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