Duality pairing in Banach spaces
$begingroup$
Suppose we have a Banach space $X$ and its dual $X^{'}$. Then the duality pairing is often written
$$langle f,vrangle_{X^{'},X} = f(v)$$
for $f in X^{'}$ and $v in X$. But is it allowed to write
$$
langle v,frangle_{X,X^{'}}
$$
to mean $langle f,vrangle_{X^{'},X}$?
In the setting of a Hilbert space $H$, for a given linear operator $A:Hto H^{'}$, I saw $A$ being self-adjont is written as
$$
langle Av,urangle = langle v,Aurangle
$$
Does $langle v,Aurangle$ here actually mean $langle Au,vrangle= Au(v)$?
functional-analysis
edited Jan 17 at 18:54
SvanN
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
$endgroup$
add a comment |
$begingroup$
Suppose we have a Banach space $X$ and its dual $X^{'}$. Then the duality pairing is often written
$$langle f,vrangle_{X^{'},X} = f(v)$$
for $f in X^{'}$ and $v in X$. But is it allowed to write
$$
langle v,frangle_{X,X^{'}}
$$
to mean $langle f,vrangle_{X^{'},X}$?
In the setting of a Hilbert space $H$, for a given linear operator $A:Hto H^{'}$, I saw $A$ being self-adjont is written as
$$
langle Av,urangle = langle v,Aurangle
$$
Does $langle v,Aurangle$ here actually mean $langle Au,vrangle= Au(v)$?
functional-analysis
edited Jan 17 at 18:54
SvanN
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
$endgroup$
$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03
add a comment |
$begingroup$
Suppose we have a Banach space $X$ and its dual $X^{'}$. Then the duality pairing is often written
$$langle f,vrangle_{X^{'},X} = f(v)$$
for $f in X^{'}$ and $v in X$. But is it allowed to write
$$
langle v,frangle_{X,X^{'}}
$$
to mean $langle f,vrangle_{X^{'},X}$?
In the setting of a Hilbert space $H$, for a given linear operator $A:Hto H^{'}$, I saw $A$ being self-adjont is written as
$$
langle Av,urangle = langle v,Aurangle
$$
Does $langle v,Aurangle$ here actually mean $langle Au,vrangle= Au(v)$?
functional-analysis
edited Jan 17 at 18:54
SvanN
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
$endgroup$
Suppose we have a Banach space $X$ and its dual $X^{'}$. Then the duality pairing is often written
$$langle f,vrangle_{X^{'},X} = f(v)$$
for $f in X^{'}$ and $v in X$. But is it allowed to write
$$
langle v,frangle_{X,X^{'}}
$$
to mean $langle f,vrangle_{X^{'},X}$?
In the setting of a Hilbert space $H$, for a given linear operator $A:Hto H^{'}$, I saw $A$ being self-adjont is written as
$$
langle Av,urangle = langle v,Aurangle
$$
Does $langle v,Aurangle$ here actually mean $langle Au,vrangle= Au(v)$?
functional-analysis
functional-analysis
edited Jan 17 at 18:54
SvanN
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
edited Jan 17 at 18:54
SvanN
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
edited Jan 17 at 18:54
SvanN
2,0411422
edited Jan 17 at 18:54
SvanN
2,0411422
edited Jan 17 at 18:54
SvanN
2,0411422
2,0411422
asked Jan 17 at 18:51
mmnnmmnn
566
asked Jan 17 at 18:51
mmnnmmnn
566
asked Jan 17 at 18:51
mmnnmmnn
566
566
$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03
add a comment |
$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03
$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03
$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03
add a comment |
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$begingroup$
It's just a convention, there's no mathematics reason why you can only write $langle v,frangle$ but not $langle f,vrangle$ and vise versa. You just need to be consistence.
$endgroup$
– BigbearZzz
Jan 17 at 19:03