Let $X$ be a Banach Space.Prove that, $X$ is strictly convex iff every points of $S(X)$ is exposed points of $B(X)$. Some definitions-


*

*$X$ is said to be strictly convex or rotund if for all $x,y\in S(X),\ x\ne y$ we have $\left\lVert\frac{x+y}{2}\right\rVert<1$

*A point $x_0\in B(X)$ is said to be a exposed point of $B(X)$ if $\exists f\in X^*$ such that $f(x_0)>f(x)\ \forall x\in B(X),\ x\ne x_0$

Here $S(X)=\{x\in X: \lVert x\rVert=1\}$ and $B(X)=\{x\in X: \lVert x\rVert\le 1\}$.
We are asked to prove the followings are equivalent-

*

*$X$ is rotund.


*All points of $S(X)$ are exposed points of $B(X)$.
I've proved $(2)\implies (1)$. If $X$ is not rotund, then $\exists x,y\in S(X),\ x\ne y$ such that $\frac{x+y}{2}\in S(X)$, hence $\frac{x+y}{2}$ is exposed point of $B(X)$. This implies $\exists f\in X^*$ such that $f\left(\frac{x+y}{2}\right)>f(z)\ \forall z\in B(X), z\ne \frac{x+y}{2}$.
In particular, $f\left(\frac{x+y}{2}\right)>f(x)\implies f(y)>f(x)$ and $f\left(\frac{x+y}{2}\right)>f(y)\implies f(x)>f(y)$, contradiction.
But I'm unable to prove the converse. I tried to do it using Hahn-Banach, Separation Theorem but not getting it. Can anyone help me in this regard? Thanks for your help in advance.
 A: Suppose $X$ is rotund. Take any $x_0\in S(X)$. By the Hahn–Banach theorem—see Theorem 5.8(b) in Folland (1999, p. 159) for details—there exists some $f\in X^*$ such that $\|f\|=1$ and $f(x_0)=\|x_0\|=1$. For the sake of contradiction, assume that $x\in B(X)\setminus\{x_0\}$ satisfies $f(x)\geq f(x_0)=1$. Clearly, $x\neq 0$, so one can define $y\equiv x/\|x\|\in S(X)$. Then,
\begin{align*}
1&=\frac{1}{2}+\frac{1}{2}\leq\frac{1}{2}+\frac{1}{2\|x\|}\leq\frac{1}{2}+\frac{f(x)}{2\|x\|}=\frac{1}{2}{f(x_0)}+\frac{1}{2}f(y)\\&=f\left(\frac{x_0+y}{2}\right)=\left|f\left(\frac{x_0+y}{2}\right)\right|\leq\|f\|\left\|\frac{x_0+y}{2}\right\|=\left\|\frac{x_0+y}{2}\right\|<1,
\end{align*}
a contradiction. The last strict inequality exploits that $x_0\neq y$. To see this, suppose for the sake of another contradiction that $x_0=y$. Then, $1=f(x_0)=f(y)=f(x/\|x\|)$, so that $f(x)=\|x\|$. But since $f(x)\geq 1$ and $\|x\|\leq 1$, this implies that $\|x\|=1$. Hence, $x_0=y=x$, which contradicts $x_0\neq x$.

Note that this proof works in any normed vector space over $\mathbb R$, not just Banach ones.
