Mixing
Rudin's functional analysis theorem 3.10, proof that multiplication is continuous
$begingroup$
Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $tau'$ makes $X$ into a locally convex space whose dual space is $X'$.
Summarizing the proof defines
$$
V = left{x : left| Lambda_i x right| < r_i, 1 leq i leq n right}
$$
If $Lambda_1, ldots, Lambda_n in X'$ and $r_i > 0$ for $i = 1,ldots, n$.
$V$ is convex, balanced and $V in tau'$, the core of the proof is to prove that the collection of all $V$ forms a local base.
For the multiplication specifically we have
Suppose $x in X$ and $alpha$ is a scalar. Then $x in sV$ for some $s>0$. If $left|beta - alpha right| < r$ and $y - x in rV$ then
$$
beta y - alpha x = (beta - alpha)y + alpha(y - x)
$$
lies in $V$ provided that $r$ is so small that
$$
r(s + r) + left| alpha right|r < 1
$$
Hence the scalar multiplication is continuous.
I'm missing here probably what the author is trying to prove exactly to show that the multiplication is continuous. Can you expound in detail what exactly is happening?
Update:
I suppose that for given $alpha$ scalar and $x in X$ the author wants to prove that for any $V$ such that $alpha x in V$ there's an open $U times W$, $U$ neighborhood of $alpha$ and $W$ neighborhood of $x$ such that if $beta in U$ and $y in W$ we have $beta y in V$, but I'm confused why he proves that $beta y - alpha x in V$ instead.
functional-analysis proof-explanation topological-vector-spaces weak-topology
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
$endgroup$
add a comment |
$begingroup$
Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $tau'$ makes $X$ into a locally convex space whose dual space is $X'$.
Summarizing the proof defines
$$
V = left{x : left| Lambda_i x right| < r_i, 1 leq i leq n right}
$$
If $Lambda_1, ldots, Lambda_n in X'$ and $r_i > 0$ for $i = 1,ldots, n$.
$V$ is convex, balanced and $V in tau'$, the core of the proof is to prove that the collection of all $V$ forms a local base.
For the multiplication specifically we have
Suppose $x in X$ and $alpha$ is a scalar. Then $x in sV$ for some $s>0$. If $left|beta - alpha right| < r$ and $y - x in rV$ then
$$
beta y - alpha x = (beta - alpha)y + alpha(y - x)
$$
lies in $V$ provided that $r$ is so small that
$$
r(s + r) + left| alpha right|r < 1
$$
Hence the scalar multiplication is continuous.
I'm missing here probably what the author is trying to prove exactly to show that the multiplication is continuous. Can you expound in detail what exactly is happening?
Update:
I suppose that for given $alpha$ scalar and $x in X$ the author wants to prove that for any $V$ such that $alpha x in V$ there's an open $U times W$, $U$ neighborhood of $alpha$ and $W$ neighborhood of $x$ such that if $beta in U$ and $y in W$ we have $beta y in V$, but I'm confused why he proves that $beta y - alpha x in V$ instead.
functional-analysis proof-explanation topological-vector-spaces weak-topology
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
$endgroup$
add a comment |
$begingroup$
Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $tau'$ makes $X$ into a locally convex space whose dual space is $X'$.
Summarizing the proof defines
$$
V = left{x : left| Lambda_i x right| < r_i, 1 leq i leq n right}
$$
If $Lambda_1, ldots, Lambda_n in X'$ and $r_i > 0$ for $i = 1,ldots, n$.
$V$ is convex, balanced and $V in tau'$, the core of the proof is to prove that the collection of all $V$ forms a local base.
For the multiplication specifically we have
Suppose $x in X$ and $alpha$ is a scalar. Then $x in sV$ for some $s>0$. If $left|beta - alpha right| < r$ and $y - x in rV$ then
$$
beta y - alpha x = (beta - alpha)y + alpha(y - x)
$$
lies in $V$ provided that $r$ is so small that
$$
r(s + r) + left| alpha right|r < 1
$$
Hence the scalar multiplication is continuous.
I'm missing here probably what the author is trying to prove exactly to show that the multiplication is continuous. Can you expound in detail what exactly is happening?
Update:
I suppose that for given $alpha$ scalar and $x in X$ the author wants to prove that for any $V$ such that $alpha x in V$ there's an open $U times W$, $U$ neighborhood of $alpha$ and $W$ neighborhood of $x$ such that if $beta in U$ and $y in W$ we have $beta y in V$, but I'm confused why he proves that $beta y - alpha x in V$ instead.
functional-analysis proof-explanation topological-vector-spaces weak-topology
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
$endgroup$
Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $tau'$ makes $X$ into a locally convex space whose dual space is $X'$.
Summarizing the proof defines
$$
V = left{x : left| Lambda_i x right| < r_i, 1 leq i leq n right}
$$
If $Lambda_1, ldots, Lambda_n in X'$ and $r_i > 0$ for $i = 1,ldots, n$.
$V$ is convex, balanced and $V in tau'$, the core of the proof is to prove that the collection of all $V$ forms a local base.
For the multiplication specifically we have
Suppose $x in X$ and $alpha$ is a scalar. Then $x in sV$ for some $s>0$. If $left|beta - alpha right| < r$ and $y - x in rV$ then
$$
beta y - alpha x = (beta - alpha)y + alpha(y - x)
$$
lies in $V$ provided that $r$ is so small that
$$
r(s + r) + left| alpha right|r < 1
$$
Hence the scalar multiplication is continuous.
I'm missing here probably what the author is trying to prove exactly to show that the multiplication is continuous. Can you expound in detail what exactly is happening?
Update:
I suppose that for given $alpha$ scalar and $x in X$ the author wants to prove that for any $V$ such that $alpha x in V$ there's an open $U times W$, $U$ neighborhood of $alpha$ and $W$ neighborhood of $x$ such that if $beta in U$ and $y in W$ we have $beta y in V$, but I'm confused why he proves that $beta y - alpha x in V$ instead.
functional-analysis proof-explanation topological-vector-spaces weak-topology
functional-analysis proof-explanation topological-vector-spaces weak-topology
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
edited Jan 9 at 17:34
user8469759
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
asked Jan 9 at 16:28
user8469759user8469759
1,4081618
1,4081618
add a comment |
add a comment |
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