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$F(y) = F(x)$ for aribtrary continuous linear functional $F$, then by Hahn-Banach $y=x$?
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I am reading an operator theory book and I frequently see something like the following used. For $x$ and $y$ in some Banach space $X$, and $F$ an arbitrary continuous linear functional on $X$, if we have that
$$
F(y) = F(x)
$$
then by the Hahn-Banach theorem $y = x$.
The Hahn-Banach theorem I know of involves extending a linear functional dominated by some sublinear function, which does not seem to come into the above statement at all. So what Hahn-Banach theorem is being used above?
functional-analysis operator-theory
asked Jul 15 at 8:25
csss
1,22811221
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up vote
1
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I am reading an operator theory book and I frequently see something like the following used. For $x$ and $y$ in some Banach space $X$, and $F$ an arbitrary continuous linear functional on $X$, if we have that
$$
F(y) = F(x)
$$
then by the Hahn-Banach theorem $y = x$.
The Hahn-Banach theorem I know of involves extending a linear functional dominated by some sublinear function, which does not seem to come into the above statement at all. So what Hahn-Banach theorem is being used above?
functional-analysis operator-theory
asked Jul 15 at 8:25
csss
1,22811221
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading an operator theory book and I frequently see something like the following used. For $x$ and $y$ in some Banach space $X$, and $F$ an arbitrary continuous linear functional on $X$, if we have that
$$
F(y) = F(x)
$$
then by the Hahn-Banach theorem $y = x$.
The Hahn-Banach theorem I know of involves extending a linear functional dominated by some sublinear function, which does not seem to come into the above statement at all. So what Hahn-Banach theorem is being used above?
functional-analysis operator-theory
asked Jul 15 at 8:25
csss
1,22811221
I am reading an operator theory book and I frequently see something like the following used. For $x$ and $y$ in some Banach space $X$, and $F$ an arbitrary continuous linear functional on $X$, if we have that
$$
F(y) = F(x)
$$
then by the Hahn-Banach theorem $y = x$.
The Hahn-Banach theorem I know of involves extending a linear functional dominated by some sublinear function, which does not seem to come into the above statement at all. So what Hahn-Banach theorem is being used above?
functional-analysis operator-theory
asked Jul 15 at 8:25
csss
1,22811221
asked Jul 15 at 8:25
csss
1,22811221
asked Jul 15 at 8:25
csss
1,22811221
asked Jul 15 at 8:25
csss
1,22811221
1,22811221
add a comment |Â
add a comment |Â
3 Answers
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Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x neq y$. Try to construct $F in operatornamespanx,y^*$ such that $F(x) neq F(y)$ - it might help to consider separately the cases where $x,y$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) neq F(y)$, contradicting your original assumption.
answered Jul 15 at 8:30
Rhys Steele
5,6101828
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up vote
1
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There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,
- The Hahn-Banach theorem (of course),
- The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c in C$ and $d in D$.
- A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $alpha$ exist such that $f(c) < alpha le f(d)$ for all $c in C$ and $d in D$.
- If $Y$ is a subspace of $X$ and $x in X$, then there exists a functional such that $Y subseteq operatornameker(f)$, and $|f| = f(x) = d_Y(x)$, where $d_Y(x) = inf_y in Y |x - y|$,
and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $lbrace x rbrace$ and $lbrace y rbrace$ quickly yields the answer.
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
add a comment |Â
up vote
0
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If $xneq y$, let $z=x-y$. Then $zneq0$. Consider the linear map $varphicolonmathbbRzlongrightarrowmathbb R$ defined by $varphi(alpha z)=alpha$. You can extend it to a continuous linear map $Fcolon Xlongrightarrowmathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1neq0$ and therefore $F(x)neq F(y)$.
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x neq y$. Try to construct $F in operatornamespanx,y^*$ such that $F(x) neq F(y)$ - it might help to consider separately the cases where $x,y$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) neq F(y)$, contradicting your original assumption.
answered Jul 15 at 8:30
Rhys Steele
5,6101828
add a comment |Â
up vote
2
down vote
Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x neq y$. Try to construct $F in operatornamespanx,y^*$ such that $F(x) neq F(y)$ - it might help to consider separately the cases where $x,y$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) neq F(y)$, contradicting your original assumption.
answered Jul 15 at 8:30
Rhys Steele
5,6101828
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x neq y$. Try to construct $F in operatornamespanx,y^*$ such that $F(x) neq F(y)$ - it might help to consider separately the cases where $x,y$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) neq F(y)$, contradicting your original assumption.
answered Jul 15 at 8:30
Rhys Steele
5,6101828
Often authors in functional analysis books will say "by Hahn-Banach" when they mean "by an easy application of or by one of the standard corollaries of Hahn-Banach" rather than "Directly from the statement of Hahn-Banach".
In this case, the Hahn-Banach theorem (as you state it) gives the result in a very straightforward way. Assume, for a contradiction, that $x neq y$. Try to construct $F in operatornamespanx,y^*$ such that $F(x) neq F(y)$ - it might help to consider separately the cases where $x,y$ is a linearly independent set and when it is not.
Once you have such a functional, by Hahn-Banach you can extend it to an element of $X^*$ such that $F(x) neq F(y)$, contradicting your original assumption.
answered Jul 15 at 8:30
Rhys Steele
5,6101828
answered Jul 15 at 8:30
Rhys Steele
5,6101828
answered Jul 15 at 8:30
Rhys Steele
5,6101828
answered Jul 15 at 8:30
Rhys Steele
5,6101828
5,6101828
add a comment |Â
add a comment |Â
up vote
1
down vote
There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,
- The Hahn-Banach theorem (of course),
- The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c in C$ and $d in D$.
- A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $alpha$ exist such that $f(c) < alpha le f(d)$ for all $c in C$ and $d in D$.
- If $Y$ is a subspace of $X$ and $x in X$, then there exists a functional such that $Y subseteq operatornameker(f)$, and $|f| = f(x) = d_Y(x)$, where $d_Y(x) = inf_y in Y |x - y|$,
and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $lbrace x rbrace$ and $lbrace y rbrace$ quickly yields the answer.
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
add a comment |Â
up vote
1
down vote
There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,
- The Hahn-Banach theorem (of course),
- The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c in C$ and $d in D$.
- A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $alpha$ exist such that $f(c) < alpha le f(d)$ for all $c in C$ and $d in D$.
- If $Y$ is a subspace of $X$ and $x in X$, then there exists a functional such that $Y subseteq operatornameker(f)$, and $|f| = f(x) = d_Y(x)$, where $d_Y(x) = inf_y in Y |x - y|$,
and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $lbrace x rbrace$ and $lbrace y rbrace$ quickly yields the answer.
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,
- The Hahn-Banach theorem (of course),
- The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c in C$ and $d in D$.
- A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $alpha$ exist such that $f(c) < alpha le f(d)$ for all $c in C$ and $d in D$.
- If $Y$ is a subspace of $X$ and $x in X$, then there exists a functional such that $Y subseteq operatornameker(f)$, and $|f| = f(x) = d_Y(x)$, where $d_Y(x) = inf_y in Y |x - y|$,
and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $lbrace x rbrace$ and $lbrace y rbrace$ quickly yields the answer.
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
There's a slew of theorems, often all referred to as "the Hahn-Banach Theorem", that are all equivalent to, or immediate consequences of, the classic Hahn-Banach theorem involving extending linear functionals dominated by sublinear functionals. Included in this are,
- The Hahn-Banach theorem (of course),
- The Hahn-Banach separation theorem: If $C$ and $D$ are convex, disjoint, and $C$ is open, then some functional $f$ exists such that $f(c) < f(d)$ for all $c in C$ and $d in D$.
- A corollary of the above using compact sets: if $C$ and $D$ are convex, and $C$ is compact, then some functional $f$ and some real $alpha$ exist such that $f(c) < alpha le f(d)$ for all $c in C$ and $d in D$.
- If $Y$ is a subspace of $X$ and $x in X$, then there exists a functional such that $Y subseteq operatornameker(f)$, and $|f| = f(x) = d_Y(x)$, where $d_Y(x) = inf_y in Y |x - y|$,
and probably others I'm forgetting. I've always preferred the geometric ones, like $2$ and $3$. Applying $3$ to the compact convex sets $lbrace x rbrace$ and $lbrace y rbrace$ quickly yields the answer.
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
edited Jul 15 at 8:42
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
answered Jul 15 at 8:36
Theo Bendit
12.1k1844
12.1k1844
add a comment |Â
add a comment |Â
up vote
0
down vote
If $xneq y$, let $z=x-y$. Then $zneq0$. Consider the linear map $varphicolonmathbbRzlongrightarrowmathbb R$ defined by $varphi(alpha z)=alpha$. You can extend it to a continuous linear map $Fcolon Xlongrightarrowmathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1neq0$ and therefore $F(x)neq F(y)$.
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
If $xneq y$, let $z=x-y$. Then $zneq0$. Consider the linear map $varphicolonmathbbRzlongrightarrowmathbb R$ defined by $varphi(alpha z)=alpha$. You can extend it to a continuous linear map $Fcolon Xlongrightarrowmathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1neq0$ and therefore $F(x)neq F(y)$.
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $xneq y$, let $z=x-y$. Then $zneq0$. Consider the linear map $varphicolonmathbbRzlongrightarrowmathbb R$ defined by $varphi(alpha z)=alpha$. You can extend it to a continuous linear map $Fcolon Xlongrightarrowmathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1neq0$ and therefore $F(x)neq F(y)$.
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
If $xneq y$, let $z=x-y$. Then $zneq0$. Consider the linear map $varphicolonmathbbRzlongrightarrowmathbb R$ defined by $varphi(alpha z)=alpha$. You can extend it to a continuous linear map $Fcolon Xlongrightarrowmathbb R$ (by the Hahn-Banach theorem). Then $F(z)=1neq0$ and therefore $F(x)neq F(y)$.
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
answered Jul 15 at 8:32
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
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