Mixing
Dualilty map in Banach space
$begingroup$
Let $X$ be a Banach space, we define
$ phi_-(y)=lim_{trightarrow{0^+}}=frac{|x|-|x-ty|}{t} $
$phi_+(y)=lim_{trightarrow{0^+}}=frac{|x+ty|-|x|}{t} $
Then $ M^*(x)= { x* in X^*: phi_-(y)leq x^*(y) leq phi_+(y) }={x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
I don't know how to prove the last equality.
banach-spaces dual-spaces
asked Jan 21 at 11:37
MikeypolMikeypol
83
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space, we define
$ phi_-(y)=lim_{trightarrow{0^+}}=frac{|x|-|x-ty|}{t} $
$phi_+(y)=lim_{trightarrow{0^+}}=frac{|x+ty|-|x|}{t} $
Then $ M^*(x)= { x* in X^*: phi_-(y)leq x^*(y) leq phi_+(y) }={x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
I don't know how to prove the last equality.
banach-spaces dual-spaces
asked Jan 21 at 11:37
MikeypolMikeypol
83
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space, we define
$ phi_-(y)=lim_{trightarrow{0^+}}=frac{|x|-|x-ty|}{t} $
$phi_+(y)=lim_{trightarrow{0^+}}=frac{|x+ty|-|x|}{t} $
Then $ M^*(x)= { x* in X^*: phi_-(y)leq x^*(y) leq phi_+(y) }={x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
I don't know how to prove the last equality.
banach-spaces dual-spaces
asked Jan 21 at 11:37
MikeypolMikeypol
83
$endgroup$
Let $X$ be a Banach space, we define
$ phi_-(y)=lim_{trightarrow{0^+}}=frac{|x|-|x-ty|}{t} $
$phi_+(y)=lim_{trightarrow{0^+}}=frac{|x+ty|-|x|}{t} $
Then $ M^*(x)= { x* in X^*: phi_-(y)leq x^*(y) leq phi_+(y) }={x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
I don't know how to prove the last equality.
banach-spaces dual-spaces
banach-spaces dual-spaces
asked Jan 21 at 11:37
MikeypolMikeypol
83
asked Jan 21 at 11:37
MikeypolMikeypol
83
asked Jan 21 at 11:37
MikeypolMikeypol
83
asked Jan 21 at 11:37
MikeypolMikeypol
83
asked Jan 21 at 11:37
MikeypolMikeypol
83
83
add a comment |
add a comment |
2 Answers
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$begingroup$
In the definition of $M^{*}(x)$ 'for all $y$' is misssing. Suppose $phi_{-} (y) leq x^{*} (y) leq phi_{+} (y)$ for all $y$. Then, taking $y=x$ we get $lim frac {1- |1-t|} {t} |x| leq x^{*} (y) leq lim frac {|1+t|-1} {t} |x|$ form which we get $|x| leq x^{*}(x) leq |x|$. Hence $x^{*} (x)=|x|$. Also, $x^{*} (y) leq phi_{+} (y) leq |y|$ for all $y$ so $|x^{*}|leq 1$. Together with $x^{*} (x)=|x|$ this implies $|x^{*}|=1$. Conversely, suppose $x^{*} (x)=|x|$ and $|x^{*}|=1$. Then $frac {|x|-|x-ty|} {t} leq frac {x^{*} (x)-x^{*} (x-ty)} t =x^{*} (y)$. Similarly, we get the second inequality.
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
add a comment |
$begingroup$
Something went wrong ! For $y=0$ we have $phi_{-}(0)=0=phi_+(0).$
Hence $M^*(x)=X^* ne {x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
answered Jan 21 at 11:56
FredFred
47.7k1849
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
In the definition of $M^{*}(x)$ 'for all $y$' is misssing. Suppose $phi_{-} (y) leq x^{*} (y) leq phi_{+} (y)$ for all $y$. Then, taking $y=x$ we get $lim frac {1- |1-t|} {t} |x| leq x^{*} (y) leq lim frac {|1+t|-1} {t} |x|$ form which we get $|x| leq x^{*}(x) leq |x|$. Hence $x^{*} (x)=|x|$. Also, $x^{*} (y) leq phi_{+} (y) leq |y|$ for all $y$ so $|x^{*}|leq 1$. Together with $x^{*} (x)=|x|$ this implies $|x^{*}|=1$. Conversely, suppose $x^{*} (x)=|x|$ and $|x^{*}|=1$. Then $frac {|x|-|x-ty|} {t} leq frac {x^{*} (x)-x^{*} (x-ty)} t =x^{*} (y)$. Similarly, we get the second inequality.
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
add a comment |
$begingroup$
In the definition of $M^{*}(x)$ 'for all $y$' is misssing. Suppose $phi_{-} (y) leq x^{*} (y) leq phi_{+} (y)$ for all $y$. Then, taking $y=x$ we get $lim frac {1- |1-t|} {t} |x| leq x^{*} (y) leq lim frac {|1+t|-1} {t} |x|$ form which we get $|x| leq x^{*}(x) leq |x|$. Hence $x^{*} (x)=|x|$. Also, $x^{*} (y) leq phi_{+} (y) leq |y|$ for all $y$ so $|x^{*}|leq 1$. Together with $x^{*} (x)=|x|$ this implies $|x^{*}|=1$. Conversely, suppose $x^{*} (x)=|x|$ and $|x^{*}|=1$. Then $frac {|x|-|x-ty|} {t} leq frac {x^{*} (x)-x^{*} (x-ty)} t =x^{*} (y)$. Similarly, we get the second inequality.
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
add a comment |
$begingroup$
In the definition of $M^{*}(x)$ 'for all $y$' is misssing. Suppose $phi_{-} (y) leq x^{*} (y) leq phi_{+} (y)$ for all $y$. Then, taking $y=x$ we get $lim frac {1- |1-t|} {t} |x| leq x^{*} (y) leq lim frac {|1+t|-1} {t} |x|$ form which we get $|x| leq x^{*}(x) leq |x|$. Hence $x^{*} (x)=|x|$. Also, $x^{*} (y) leq phi_{+} (y) leq |y|$ for all $y$ so $|x^{*}|leq 1$. Together with $x^{*} (x)=|x|$ this implies $|x^{*}|=1$. Conversely, suppose $x^{*} (x)=|x|$ and $|x^{*}|=1$. Then $frac {|x|-|x-ty|} {t} leq frac {x^{*} (x)-x^{*} (x-ty)} t =x^{*} (y)$. Similarly, we get the second inequality.
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
$endgroup$
In the definition of $M^{*}(x)$ 'for all $y$' is misssing. Suppose $phi_{-} (y) leq x^{*} (y) leq phi_{+} (y)$ for all $y$. Then, taking $y=x$ we get $lim frac {1- |1-t|} {t} |x| leq x^{*} (y) leq lim frac {|1+t|-1} {t} |x|$ form which we get $|x| leq x^{*}(x) leq |x|$. Hence $x^{*} (x)=|x|$. Also, $x^{*} (y) leq phi_{+} (y) leq |y|$ for all $y$ so $|x^{*}|leq 1$. Together with $x^{*} (x)=|x|$ this implies $|x^{*}|=1$. Conversely, suppose $x^{*} (x)=|x|$ and $|x^{*}|=1$. Then $frac {|x|-|x-ty|} {t} leq frac {x^{*} (x)-x^{*} (x-ty)} t =x^{*} (y)$. Similarly, we get the second inequality.
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
answered Jan 21 at 11:57
Kavi Rama MurthyKavi Rama Murthy
65.3k42766
65.3k42766
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
add a comment |
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
$begingroup$
Thank you so much.
$endgroup$
– Mikeypol
Jan 21 at 12:22
add a comment |
$begingroup$
Something went wrong ! For $y=0$ we have $phi_{-}(0)=0=phi_+(0).$
Hence $M^*(x)=X^* ne {x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
answered Jan 21 at 11:56
FredFred
47.7k1849
$endgroup$
add a comment |
$begingroup$
Something went wrong ! For $y=0$ we have $phi_{-}(0)=0=phi_+(0).$
Hence $M^*(x)=X^* ne {x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
answered Jan 21 at 11:56
FredFred
47.7k1849
$endgroup$
add a comment |
$begingroup$
Something went wrong ! For $y=0$ we have $phi_{-}(0)=0=phi_+(0).$
Hence $M^*(x)=X^* ne {x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
answered Jan 21 at 11:56
FredFred
47.7k1849
$endgroup$
Something went wrong ! For $y=0$ we have $phi_{-}(0)=0=phi_+(0).$
Hence $M^*(x)=X^* ne {x^* in X^*: |x^*|=1 and x*(x)=|x|}.$
answered Jan 21 at 11:56
FredFred
47.7k1849
answered Jan 21 at 11:56
FredFred
47.7k1849
answered Jan 21 at 11:56
FredFred
47.7k1849
answered Jan 21 at 11:56
FredFred
47.7k1849
47.7k1849
add a comment |
add a comment |
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