weak* topology is not defined by any translation invariant metric when $X$ is infinite dimensional
$begingroup$
There is an exercise in Folland's real analysis, page 170
If $X$ is an infinite dimensional Banach space, then the weak* topology is not defined by any translation-invariant metric.
He gives a hint to this problem saying that
Every weak*-Cauchy sequence in $X^*$ converges.
I don't know how to use his hint and the condition on "translation-invariance" to approach this problem. I know I should use the unboundedness of weak-open and weak*-open sets somewhere in the proof.
Furthermore, I wonder if the weak* topology can be defined by some metric that is not translation-invariant, which is a weaker result for this problem.
Thanks in advance.
functional-analysis
asked May 3 '15 at 2:59
No OneNo One
2,1151519
$endgroup$
add a comment |
$begingroup$
There is an exercise in Folland's real analysis, page 170
If $X$ is an infinite dimensional Banach space, then the weak* topology is not defined by any translation-invariant metric.
He gives a hint to this problem saying that
Every weak*-Cauchy sequence in $X^*$ converges.
I don't know how to use his hint and the condition on "translation-invariance" to approach this problem. I know I should use the unboundedness of weak-open and weak*-open sets somewhere in the proof.
Furthermore, I wonder if the weak* topology can be defined by some metric that is not translation-invariant, which is a weaker result for this problem.
Thanks in advance.
functional-analysis
asked May 3 '15 at 2:59
No OneNo One
2,1151519
$endgroup$
$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07
add a comment |
$begingroup$
There is an exercise in Folland's real analysis, page 170
If $X$ is an infinite dimensional Banach space, then the weak* topology is not defined by any translation-invariant metric.
He gives a hint to this problem saying that
Every weak*-Cauchy sequence in $X^*$ converges.
I don't know how to use his hint and the condition on "translation-invariance" to approach this problem. I know I should use the unboundedness of weak-open and weak*-open sets somewhere in the proof.
Furthermore, I wonder if the weak* topology can be defined by some metric that is not translation-invariant, which is a weaker result for this problem.
Thanks in advance.
functional-analysis
asked May 3 '15 at 2:59
No OneNo One
2,1151519
$endgroup$
There is an exercise in Folland's real analysis, page 170
If $X$ is an infinite dimensional Banach space, then the weak* topology is not defined by any translation-invariant metric.
He gives a hint to this problem saying that
Every weak*-Cauchy sequence in $X^*$ converges.
I don't know how to use his hint and the condition on "translation-invariance" to approach this problem. I know I should use the unboundedness of weak-open and weak*-open sets somewhere in the proof.
Furthermore, I wonder if the weak* topology can be defined by some metric that is not translation-invariant, which is a weaker result for this problem.
Thanks in advance.
functional-analysis
functional-analysis
asked May 3 '15 at 2:59
No OneNo One
2,1151519
asked May 3 '15 at 2:59
No OneNo One
2,1151519
asked May 3 '15 at 2:59
No OneNo One
2,1151519
asked May 3 '15 at 2:59
No OneNo One
2,1151519
asked May 3 '15 at 2:59
No OneNo One
2,1151519
2,1151519
$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07
add a comment |
$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07
$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07
$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07
add a comment |
1 Answer
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$begingroup$
Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:
Every nonempty weak$^*$-open set in $X^*$ is unbounded.
I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 in X^*$ and for $n=1,2,dots$, pick $f_n $ in the weak$^*$-open ball ${f:d(f,f_0)<1/n}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then ${f_n}$ is weak$^*$-Cauchy because given any $epsilon > 0$, we can pick $N$ such that $2/N < epsilon$, whence $d(f_n,f_m) le d(f_n, f_0)+f(f_0, f_m) <2/N <epsilon$ whenever $m,n ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)to f(x)$ for all $xin X$. This contradicts the Uniform Boundedness Principle since we have $sup_n |f_n(x)| < infty$ for all $xin X$ and yet $sup_n || f_n|| = infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
answered Jan 19 at 5:23
suncup224suncup224
1,211816
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add a comment |
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$begingroup$
Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:
Every nonempty weak$^*$-open set in $X^*$ is unbounded.
I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 in X^*$ and for $n=1,2,dots$, pick $f_n $ in the weak$^*$-open ball ${f:d(f,f_0)<1/n}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then ${f_n}$ is weak$^*$-Cauchy because given any $epsilon > 0$, we can pick $N$ such that $2/N < epsilon$, whence $d(f_n,f_m) le d(f_n, f_0)+f(f_0, f_m) <2/N <epsilon$ whenever $m,n ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)to f(x)$ for all $xin X$. This contradicts the Uniform Boundedness Principle since we have $sup_n |f_n(x)| < infty$ for all $xin X$ and yet $sup_n || f_n|| = infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
answered Jan 19 at 5:23
suncup224suncup224
1,211816
$endgroup$
add a comment |
$begingroup$
Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:
Every nonempty weak$^*$-open set in $X^*$ is unbounded.
I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 in X^*$ and for $n=1,2,dots$, pick $f_n $ in the weak$^*$-open ball ${f:d(f,f_0)<1/n}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then ${f_n}$ is weak$^*$-Cauchy because given any $epsilon > 0$, we can pick $N$ such that $2/N < epsilon$, whence $d(f_n,f_m) le d(f_n, f_0)+f(f_0, f_m) <2/N <epsilon$ whenever $m,n ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)to f(x)$ for all $xin X$. This contradicts the Uniform Boundedness Principle since we have $sup_n |f_n(x)| < infty$ for all $xin X$ and yet $sup_n || f_n|| = infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
answered Jan 19 at 5:23
suncup224suncup224
1,211816
$endgroup$
add a comment |
$begingroup$
Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:
Every nonempty weak$^*$-open set in $X^*$ is unbounded.
I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 in X^*$ and for $n=1,2,dots$, pick $f_n $ in the weak$^*$-open ball ${f:d(f,f_0)<1/n}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then ${f_n}$ is weak$^*$-Cauchy because given any $epsilon > 0$, we can pick $N$ such that $2/N < epsilon$, whence $d(f_n,f_m) le d(f_n, f_0)+f(f_0, f_m) <2/N <epsilon$ whenever $m,n ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)to f(x)$ for all $xin X$. This contradicts the Uniform Boundedness Principle since we have $sup_n |f_n(x)| < infty$ for all $xin X$ and yet $sup_n || f_n|| = infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
answered Jan 19 at 5:23
suncup224suncup224
1,211816
$endgroup$
Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:
Every nonempty weak$^*$-open set in $X^*$ is unbounded.
I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 in X^*$ and for $n=1,2,dots$, pick $f_n $ in the weak$^*$-open ball ${f:d(f,f_0)<1/n}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then ${f_n}$ is weak$^*$-Cauchy because given any $epsilon > 0$, we can pick $N$ such that $2/N < epsilon$, whence $d(f_n,f_m) le d(f_n, f_0)+f(f_0, f_m) <2/N <epsilon$ whenever $m,n ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)to f(x)$ for all $xin X$. This contradicts the Uniform Boundedness Principle since we have $sup_n |f_n(x)| < infty$ for all $xin X$ and yet $sup_n || f_n|| = infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
answered Jan 19 at 5:23
suncup224suncup224
1,211816
answered Jan 19 at 5:23
suncup224suncup224
1,211816
answered Jan 19 at 5:23
suncup224suncup224
1,211816
answered Jan 19 at 5:23
suncup224suncup224
1,211816
1,211816
add a comment |
add a comment |
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$begingroup$
The weak* topology is not metrizable in any infinite dimensional vector space. One way to see this, I think, is to show that that topology is not first countable.
$endgroup$
– Mariano Suárez-Álvarez
May 3 '15 at 3:07