I read a paper "on operators which attain their norm" by J. Lindenstrauss, and have some problem in theorem 2 (p.142~143). First I give some definitions to be needed later.
Def)1. For Banach spaces X and Y, $B(X,Y)$ is a set of all bounded linear operator from X to Y.
An element T of $B(X,Y)$ attains its norm if $\Vert T\Vert=\Vert Tx\Vert$ for some $x\in X$
Let $C$ be a convex subset of a Banach space X. $x\in C$ is called an exposed point if there is $f\in X^*$, the dual of $X$, such that $f(x)>f(y) \forall y\neq x$ in C.
The situation is as follows: $X$ is a Banach space such that every element of $B(X,Y)$ attains its norm, and T is an embedding from X to Y. I wanna show if $\Vert Tx\Vert=\Vert T\Vert$, then x is an exposed point of $S_X$.
(In fact it has more condition that the range of T is locally uniformly convex, which is omitted because it is related to the strong uniform convexity only, I think)
The proof is, choose $g\in Y^*, \Vert g\Vert=1$ such that $g(Tx)=\Vert Tx\Vert=\Vert T\Vert$ (and it skipped all other things).
I know for every $y\neq x$ in $S_X$, $g(Ty)\leq \Vert g\Vert\Vert Ty\Vert\leq\Vert T\Vert=g(Tx)$, but I cannot yields the strict inequality unless $g$ is one to one, so I'm stuck here. How can I do it?
In fact, it needs the locally uniformly convex (l.u.c.) condition. Once we give this, then for $\Vert x_n\Vert\leq 1$, $g(Tx_n)\rightarrow \Vert T \Vert$ implies $\Vert Tx_n-Tx\Vert\rightarrow 0$ and hence $\Vert x_n-x \Vert\rightarrow 0$ by the injectivity of T. If $g(Tx)=g(Ty)$, take $x_n=y \forall n$ to conclude x=y.