Mixing
Linearity of functionals imply the inclusion of kernels
$begingroup$
I understand most of the following claim's proof except statement I've highlighted below:
Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.
Proof:
The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
$$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
$$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.
I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?
functional-analysis proof-explanation
edited Jan 8 at 22:58
mechanodroid
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
$endgroup$
add a comment |
$begingroup$
I understand most of the following claim's proof except statement I've highlighted below:
Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.
Proof:
The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
$$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
$$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.
I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?
functional-analysis proof-explanation
edited Jan 8 at 22:58
mechanodroid
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
$endgroup$
add a comment |
$begingroup$
I understand most of the following claim's proof except statement I've highlighted below:
Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.
Proof:
The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
$$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
$$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.
I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?
functional-analysis proof-explanation
edited Jan 8 at 22:58
mechanodroid
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
$endgroup$
I understand most of the following claim's proof except statement I've highlighted below:
Let $X$ be a linear space and $W$ be a subspace of $X^{text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.
Proof:
The reverse directions follows from definition. In the forward direction, suppose $psi:X rightarrow mathbb{R}$ is $W$-weakly continuous. By the continuity of $psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|psi(x)| = |psi(x)-psi(0)| < 1$ if $x in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{epsilon, psi_1, ldots, psi_n}$) at $0$ that is contained in $N$. Choose $epsilon > 0$ and $psi_1,...psi_n$ in $W$ for which:
$$N_{epsilon, psi_1, ldots, psi_n}= {x' in X ,big| ,, |psi_k(x')-psi_k(0)| < epsilon quad 1 leq k leq n} subset N$$Thus:
$$|psi(x)|<1 text{ if } |psi_k(x)|<epsilon quad forall 1leq k leq n$$
By linearity of $psi$ and $psi_k$ we have the inclusion $bigcap ker psi_k subset ker psi$. By a previous theorem, this implies $psi$ is a linear combination of the $psi_k$'s.
I don't understand the bolded statement. Namely, suppose $A subset N_{epsilon, psi_1, ldots, psi_n}$ is the set of all $x$ for which all the $psi_k$ are zero -- i.e. $A = bigcap ker psi_k$. The implication shows $|psi(A)|<1$. This doesn't imply $|psi(A)|=0$. So how do we get $A subset ker psi$?
functional-analysis proof-explanation
functional-analysis proof-explanation
edited Jan 8 at 22:58
mechanodroid
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
edited Jan 8 at 22:58
mechanodroid
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
edited Jan 8 at 22:58
mechanodroid
27.3k62446
edited Jan 8 at 22:58
mechanodroid
27.3k62446
edited Jan 8 at 22:58
mechanodroid
27.3k62446
27.3k62446
asked Jan 8 at 21:56
yoshiyoshi
1,191817
asked Jan 8 at 21:56
yoshiyoshi
1,191817
asked Jan 8 at 21:56
yoshiyoshi
1,191817
1,191817
add a comment |
add a comment |
1 Answer
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Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.
But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
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add a comment |
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$begingroup$
Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.
But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.
But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.
But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
$endgroup$
Assume that $x in bigcap_{k=1}^n ker psi_k$. Then $|psi_k(x)| = 0 < varepsilon$ for all $1 le k le n$ so it follows $|psi(x)| < 1$.
But also $n x in bigcap_{k=1}^n ker psi_k$ for any $n in mathbb{N}$ so similarly $n|psi(x)| = |psi(nx)| < 1$. If $psi(x) ne 0$, the left hand side would be unbounded as $n to infty$ so it has to be $psi(x) = 0$.
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 22:56
mechanodroidmechanodroid
27.3k62446
27.3k62446
add a comment |
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