Norm and Weak Topologies agree?
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so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.
general-topology functional-analysis weak-convergence
asked 4 hours ago
ScottScott
34918
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add a comment |
$begingroup$
so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.
general-topology functional-analysis weak-convergence
asked 4 hours ago
ScottScott
34918
$endgroup$
add a comment |
$begingroup$
so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.
general-topology functional-analysis weak-convergence
asked 4 hours ago
ScottScott
34918
$endgroup$
so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_alpharightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $fin X^*$ we know by continuity that $f(x_alpha)rightarrow f(x)$, which means $x_alpharightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.
general-topology functional-analysis weak-convergence
general-topology functional-analysis weak-convergence
asked 4 hours ago
ScottScott
34918
asked 4 hours ago
ScottScott
34918
asked 4 hours ago
ScottScott
34918
asked 4 hours ago
ScottScott
34918
asked 4 hours ago
ScottScott
34918
34918
add a comment |
add a comment |
2 Answers
2
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We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.
More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
add a comment |
$begingroup$
Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$
Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$
answered 20 mins ago
MatematletaMatematleta
10.5k2918
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add a comment |
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2 Answers
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2 Answers
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active
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active
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$begingroup$
We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.
More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
add a comment |
$begingroup$
We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.
More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
$endgroup$
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
add a comment |
$begingroup$
We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.
More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
$endgroup$
We can identify our space $X$ with $mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_alphato x$ weakly, then $f_i(x_alpha)to f_i(x)$ where $f_i:mathbb{R}^ntomathbb{R}$ is the $i$th coordinate functional. But that just means that $x_alpha$ converges to $x$ on each coordinate, which implies $x_alpha$ converges to $x$ in the Euclidean topology. Thus $x_alphato x$ in the norm topology.
More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
edited 3 hours ago
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
answered 4 hours ago
Eric WofseyEric Wofsey
183k13211338
183k13211338
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
add a comment |
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
So this proof fails in the infinite dimensional case as we can't identify our space with $mathbb{R}^n$, correct?
$endgroup$
– Scott
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
$begingroup$
Yes, that's correct.
$endgroup$
– Eric Wofsey
3 hours ago
add a comment |
$begingroup$
Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$
Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$
answered 20 mins ago
MatematletaMatematleta
10.5k2918
$endgroup$
add a comment |
$begingroup$
Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$
Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$
answered 20 mins ago
MatematletaMatematleta
10.5k2918
$endgroup$
add a comment |
$begingroup$
Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$
Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$
answered 20 mins ago
MatematletaMatematleta
10.5k2918
$endgroup$
Another way to do this is to just go back to the definitions. Let $epsilon>0,$ and $x_0in X$. It suffices to show that any open ball $B(x_0,epsilon)$ is weakly open. We may use $|cdot|_{infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=sum^n_{i=1} x_ie_i$ for each $xin X$ and for each $1le ile n,$ we have the functionals $e_i^*:Xto mathbb C:xmapsto x_i.$
Then, $B_{epsilon }=left { x:left | xright |<epsilon right }=left { x:forall 1le ile n, |x_i|<epsilon right }, $ which is nothing more than the weakly open set $left { x:forall 1le ile n, |e_i^*(x)|<epsilon right }.$
answered 20 mins ago
MatematletaMatematleta
10.5k2918
edited 11 mins ago
answered 20 mins ago
MatematletaMatematleta
10.5k2918
answered 20 mins ago
MatematletaMatematleta
10.5k2918
answered 20 mins ago
MatematletaMatematleta
10.5k2918
10.5k2918
add a comment |
add a comment |
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