Dualilty map in Banach space

Question

Let $X$ be a Banach space, we define

$ \phi_-(y)=\lim_{t\rightarrow{0^+}}=\frac{|x|-|x-ty|}{t} $
$\phi_+(y)=\lim_{t\rightarrow{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.

Answer 1

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.

Answer 2

Something went wrong ! For $y=0$ we have $\phi_{-}(0)=0=\phi_+(0).$

Hence $M^*(x)=X^* \ne \{x^* \in X^*: |x^*|=1 \ and \ x*(x)=|x|\}.$