Limit of subsequence equals weak* limit
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Suppose I have a convergent sequence $T_n to^* T$, converging in the weak* topology in the dual space ${ T_n } subset X^*$ of a locally convex topological vector space $X$.
Further assume that I know the sequence is taken in some subset $C subset X^*$ which is compact with respect to a metric $d : X^* times X^* to R$, i.e. $(C, d)$ is a compact metric space.
Thus I can extract a convergent subsequence ${ T_{n_k} }$ which converges to some limit point $tilde T in C$ in $d$-metric.
Further I know that the original weak$^*$ limit $T$ is also in $C$, i.e. $T in C$.
Is it possible to prove that $tilde T = T$ and that the whole sequence ${ T_n }$ converges in metric $d$ to the weak limit $T$? I.e., I want to show $d(T_n, T) to 0$.
The background is, that I want to prove that a certain metric $d$ metrizes the weak$^*$ topology of my space on a certain subset.
functional-analysis
asked Dec 31 '18 at 13:01
yonyon
689
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add a comment |
$begingroup$
Suppose I have a convergent sequence $T_n to^* T$, converging in the weak* topology in the dual space ${ T_n } subset X^*$ of a locally convex topological vector space $X$.
Further assume that I know the sequence is taken in some subset $C subset X^*$ which is compact with respect to a metric $d : X^* times X^* to R$, i.e. $(C, d)$ is a compact metric space.
Thus I can extract a convergent subsequence ${ T_{n_k} }$ which converges to some limit point $tilde T in C$ in $d$-metric.
Further I know that the original weak$^*$ limit $T$ is also in $C$, i.e. $T in C$.
Is it possible to prove that $tilde T = T$ and that the whole sequence ${ T_n }$ converges in metric $d$ to the weak limit $T$? I.e., I want to show $d(T_n, T) to 0$.
The background is, that I want to prove that a certain metric $d$ metrizes the weak$^*$ topology of my space on a certain subset.
functional-analysis
asked Dec 31 '18 at 13:01
yonyon
689
$endgroup$
add a comment |
$begingroup$
Suppose I have a convergent sequence $T_n to^* T$, converging in the weak* topology in the dual space ${ T_n } subset X^*$ of a locally convex topological vector space $X$.
Further assume that I know the sequence is taken in some subset $C subset X^*$ which is compact with respect to a metric $d : X^* times X^* to R$, i.e. $(C, d)$ is a compact metric space.
Thus I can extract a convergent subsequence ${ T_{n_k} }$ which converges to some limit point $tilde T in C$ in $d$-metric.
Further I know that the original weak$^*$ limit $T$ is also in $C$, i.e. $T in C$.
Is it possible to prove that $tilde T = T$ and that the whole sequence ${ T_n }$ converges in metric $d$ to the weak limit $T$? I.e., I want to show $d(T_n, T) to 0$.
The background is, that I want to prove that a certain metric $d$ metrizes the weak$^*$ topology of my space on a certain subset.
functional-analysis
asked Dec 31 '18 at 13:01
yonyon
689
$endgroup$
Suppose I have a convergent sequence $T_n to^* T$, converging in the weak* topology in the dual space ${ T_n } subset X^*$ of a locally convex topological vector space $X$.
Further assume that I know the sequence is taken in some subset $C subset X^*$ which is compact with respect to a metric $d : X^* times X^* to R$, i.e. $(C, d)$ is a compact metric space.
Thus I can extract a convergent subsequence ${ T_{n_k} }$ which converges to some limit point $tilde T in C$ in $d$-metric.
Further I know that the original weak$^*$ limit $T$ is also in $C$, i.e. $T in C$.
Is it possible to prove that $tilde T = T$ and that the whole sequence ${ T_n }$ converges in metric $d$ to the weak limit $T$? I.e., I want to show $d(T_n, T) to 0$.
The background is, that I want to prove that a certain metric $d$ metrizes the weak$^*$ topology of my space on a certain subset.
functional-analysis
functional-analysis
asked Dec 31 '18 at 13:01
yonyon
689
asked Dec 31 '18 at 13:01
yonyon
689
asked Dec 31 '18 at 13:01
yonyon
689
asked Dec 31 '18 at 13:01
yonyon
689
asked Dec 31 '18 at 13:01
yonyon
689
689
add a comment |
add a comment |
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