Relation between the weak* topology and a metric topology on a separable normed linear space
Let $X$ be a normed $mathbb{K}$ -linear space and let $B^{ast}={ fin X^ast ,||f||leq 1$ },then $B^ast$ is metrisable $iff$ $X$ is separable. The corresponding metric is defined by $$d(f,g)=sum_{ninmathbb{N}}frac{1}{2^n ||x_n||}|f(x_n)-g(x_n)|$$ Now this metric also induces a topology(say $T_d$) on $X^ast$ .What is the relation between the weak* topology on $X^ast$ and$(X^ast,T_d)$?
functional-analysis
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
add a comment |
Let $X$ be a normed $mathbb{K}$ -linear space and let $B^{ast}={ fin X^ast ,||f||leq 1$ },then $B^ast$ is metrisable $iff$ $X$ is separable. The corresponding metric is defined by $$d(f,g)=sum_{ninmathbb{N}}frac{1}{2^n ||x_n||}|f(x_n)-g(x_n)|$$ Now this metric also induces a topology(say $T_d$) on $X^ast$ .What is the relation between the weak* topology on $X^ast$ and$(X^ast,T_d)$?
functional-analysis
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40
add a comment |
Let $X$ be a normed $mathbb{K}$ -linear space and let $B^{ast}={ fin X^ast ,||f||leq 1$ },then $B^ast$ is metrisable $iff$ $X$ is separable. The corresponding metric is defined by $$d(f,g)=sum_{ninmathbb{N}}frac{1}{2^n ||x_n||}|f(x_n)-g(x_n)|$$ Now this metric also induces a topology(say $T_d$) on $X^ast$ .What is the relation between the weak* topology on $X^ast$ and$(X^ast,T_d)$?
functional-analysis
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
Let $X$ be a normed $mathbb{K}$ -linear space and let $B^{ast}={ fin X^ast ,||f||leq 1$ },then $B^ast$ is metrisable $iff$ $X$ is separable. The corresponding metric is defined by $$d(f,g)=sum_{ninmathbb{N}}frac{1}{2^n ||x_n||}|f(x_n)-g(x_n)|$$ Now this metric also induces a topology(say $T_d$) on $X^ast$ .What is the relation between the weak* topology on $X^ast$ and$(X^ast,T_d)$?
functional-analysis
functional-analysis
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
edited Dec 2 '18 at 14:39
Henno Brandsma
105k347114
105k347114
asked Dec 2 '18 at 14:19
Samiron Parui
1808
asked Dec 2 '18 at 14:19
Samiron Parui
1808
asked Dec 2 '18 at 14:19
Samiron Parui
1808
1808
sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40
add a comment |
sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40
sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40
add a comment |
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sorry,it should be X*
– Samiron Parui
Dec 2 '18 at 14:36
The formula you give need not converge for arbitary $f,g$; the scaling factors works for $B^ast$, not for $X^ast$; so how are you solving that ?
– Henno Brandsma
Dec 2 '18 at 14:40