Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.
$begingroup$
Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.
Could anybody give me some hints to start?
general-topology functional-analysis topological-vector-spaces product-space weak-topology
asked Dec 11 '18 at 22:09
RossRoss
1237
$endgroup$
add a comment |
$begingroup$
Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.
Could anybody give me some hints to start?
general-topology functional-analysis topological-vector-spaces product-space weak-topology
asked Dec 11 '18 at 22:09
RossRoss
1237
$endgroup$
$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51
add a comment |
$begingroup$
Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.
Could anybody give me some hints to start?
general-topology functional-analysis topological-vector-spaces product-space weak-topology
asked Dec 11 '18 at 22:09
RossRoss
1237
$endgroup$
Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.
Could anybody give me some hints to start?
general-topology functional-analysis topological-vector-spaces product-space weak-topology
general-topology functional-analysis topological-vector-spaces product-space weak-topology
asked Dec 11 '18 at 22:09
RossRoss
1237
asked Dec 11 '18 at 22:09
RossRoss
1237
edited Dec 11 '18 at 22:42
Ross
asked Dec 11 '18 at 22:09
RossRoss
1237
asked Dec 11 '18 at 22:09
RossRoss
1237
asked Dec 11 '18 at 22:09
RossRoss
1237
1237
$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51
add a comment |
$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51
$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51
add a comment |
2 Answers
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You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.
Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
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add a comment |
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You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.
I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.
Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.
Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.
Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.
Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
61.8k42262
add a comment |
add a comment |
$begingroup$
You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.
I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.
I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.
I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
$endgroup$
You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.
I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
answered Dec 11 '18 at 23:20
MindlackMindlack
4,740210
4,740210
add a comment |
add a comment |
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$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16
$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41
$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51