Mixing
Understanding the defintion of dual operators
$begingroup$
I'm reading a book about Functional Analysis, and now I've reached to the part about dual operators.
I'm having some difficulties understanding the following definition -
Why $A^*$ is $Y^*rightarrow X^*$?
We know that $phi in Y^*$, i.e., $phi:Yrightarrow Y$ and bounded. So the image of $phi(Ax)$ should be in $Y^*$, shouldn't it?
How come it's in $X^*$?
($X^*$ is reffered here as the space of all bounded linear functionals on $X$).
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 17 at 22:03
ChikChakChikChak
764418
$endgroup$
add a comment |
$begingroup$
I'm reading a book about Functional Analysis, and now I've reached to the part about dual operators.
I'm having some difficulties understanding the following definition -
Why $A^*$ is $Y^*rightarrow X^*$?
We know that $phi in Y^*$, i.e., $phi:Yrightarrow Y$ and bounded. So the image of $phi(Ax)$ should be in $Y^*$, shouldn't it?
How come it's in $X^*$?
($X^*$ is reffered here as the space of all bounded linear functionals on $X$).
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 17 at 22:03
ChikChakChikChak
764418
$endgroup$
add a comment |
$begingroup$
I'm reading a book about Functional Analysis, and now I've reached to the part about dual operators.
I'm having some difficulties understanding the following definition -
Why $A^*$ is $Y^*rightarrow X^*$?
We know that $phi in Y^*$, i.e., $phi:Yrightarrow Y$ and bounded. So the image of $phi(Ax)$ should be in $Y^*$, shouldn't it?
How come it's in $X^*$?
($X^*$ is reffered here as the space of all bounded linear functionals on $X$).
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 17 at 22:03
ChikChakChikChak
764418
$endgroup$
I'm reading a book about Functional Analysis, and now I've reached to the part about dual operators.
I'm having some difficulties understanding the following definition -
Why $A^*$ is $Y^*rightarrow X^*$?
We know that $phi in Y^*$, i.e., $phi:Yrightarrow Y$ and bounded. So the image of $phi(Ax)$ should be in $Y^*$, shouldn't it?
How come it's in $X^*$?
($X^*$ is reffered here as the space of all bounded linear functionals on $X$).
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 17 at 22:03
ChikChakChikChak
764418
asked Jan 17 at 22:03
ChikChakChikChak
764418
asked Jan 17 at 22:03
ChikChakChikChak
764418
asked Jan 17 at 22:03
ChikChakChikChak
764418
asked Jan 17 at 22:03
ChikChakChikChak
764418
764418
add a comment |
add a comment |
1 Answer
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$begingroup$
First of all, $phi$ is a functional, so it is $phi:Ytomathbb{C}$.
Next, what is $A^*$? It takes a functional $phiin Y^*$ and returns a function $A^*(phi)=f$ which is defined by $f(x)=phi(A(x))$. If we take a vector $xin X$ then $A(x)in Y$ and then $phi(A(x))inmathbb{C}$. So $f:Xtomathbb{C}$. Then of course you need to check that $f$ is linear and bounded, but that's easy. So $fin X^*$.
answered Jan 17 at 22:14
MarkMark
9,409622
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1 Answer
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$begingroup$
First of all, $phi$ is a functional, so it is $phi:Ytomathbb{C}$.
Next, what is $A^*$? It takes a functional $phiin Y^*$ and returns a function $A^*(phi)=f$ which is defined by $f(x)=phi(A(x))$. If we take a vector $xin X$ then $A(x)in Y$ and then $phi(A(x))inmathbb{C}$. So $f:Xtomathbb{C}$. Then of course you need to check that $f$ is linear and bounded, but that's easy. So $fin X^*$.
answered Jan 17 at 22:14
MarkMark
9,409622
$endgroup$
add a comment |
$begingroup$
First of all, $phi$ is a functional, so it is $phi:Ytomathbb{C}$.
Next, what is $A^*$? It takes a functional $phiin Y^*$ and returns a function $A^*(phi)=f$ which is defined by $f(x)=phi(A(x))$. If we take a vector $xin X$ then $A(x)in Y$ and then $phi(A(x))inmathbb{C}$. So $f:Xtomathbb{C}$. Then of course you need to check that $f$ is linear and bounded, but that's easy. So $fin X^*$.
answered Jan 17 at 22:14
MarkMark
9,409622
$endgroup$
add a comment |
$begingroup$
First of all, $phi$ is a functional, so it is $phi:Ytomathbb{C}$.
Next, what is $A^*$? It takes a functional $phiin Y^*$ and returns a function $A^*(phi)=f$ which is defined by $f(x)=phi(A(x))$. If we take a vector $xin X$ then $A(x)in Y$ and then $phi(A(x))inmathbb{C}$. So $f:Xtomathbb{C}$. Then of course you need to check that $f$ is linear and bounded, but that's easy. So $fin X^*$.
answered Jan 17 at 22:14
MarkMark
9,409622
$endgroup$
First of all, $phi$ is a functional, so it is $phi:Ytomathbb{C}$.
Next, what is $A^*$? It takes a functional $phiin Y^*$ and returns a function $A^*(phi)=f$ which is defined by $f(x)=phi(A(x))$. If we take a vector $xin X$ then $A(x)in Y$ and then $phi(A(x))inmathbb{C}$. So $f:Xtomathbb{C}$. Then of course you need to check that $f$ is linear and bounded, but that's easy. So $fin X^*$.
answered Jan 17 at 22:14
MarkMark
9,409622
answered Jan 17 at 22:14
MarkMark
9,409622
answered Jan 17 at 22:14
MarkMark
9,409622
answered Jan 17 at 22:14
MarkMark
9,409622
9,409622
add a comment |
add a comment |
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