Relation between Frechet differentiability of $X^*$ and reflexivity of $X$

Question

Let $X$ be a Banach space and $X^*$ be dual of $X$. If $X^*$ is Frechet differentiability then $X$ is reflexive. See Theorem 8.6 in:

Fabian, M. J. (Ed.). (2001). Functional analysis and infinite-dimensional geometry (Vol. 8). Springer.

But I'd like to prove this theorem by this lemma

Lemma 8.4

Let $(X,\|.\|)$ be a Banach space, and let $x\in S_X$:

$\|.\|$ is Frechet differentiable at $x$ if and only if $\lim\limits_{n\to\infty}\|f_n-g_n\|$ whenever $f_n,g_n \in S_{X^*}$ satisfy$\lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}g_n(x)=1$ if and only if $\{f_n\}\subset S_{X^*}$ is convergent whenever$\lim\limits_{n\to\infty}f_n(x)=1.$

Any ideas would be greatly appreciated.

Answer 1

I guess you just want to know how to interpret the proof given in the textbook, since it already uses Lemma 8.4. Recall:

Corollary 3.56. A Banach space $X$ is reflexive iff every $f\in X^*$ attains its norm.

So, to prove Theorem 8.6, we just need to show that a given $f\in S_{X^*}$ (the unit sphere of $X^*$) attains its norm.

Note that the relevant part of Lemma 8.4 is this:

Lemma 8.4 (relevant part only.) If the norm $\|\cdot\|$ of a Banach space $X$ is Frechet-differentiable at $x\in S_X$ then the following condition holds: \begin{equation}\tag{1}\text{if }(f_n)_{n=1}^\infty\subset S_{X^*}\text{ and }\lim_{n\to\infty}f_n(x)=1,\text{ then }(f_n)_{n=1}^\infty\text{ is norm-convergent in }X^*.\end{equation}

By substituting $X^*$ for $X$ we get the following:

Corollary A. Let $X$ be a Banach space If the norm $\|\cdot\|$ of $X^*$ is Frechet-differentiable at $f\in S_{X^*}$ then the following condition holds: \begin{equation}\tag{2}\text{if }(x_n)_{n=1}^\infty\subset S_{X^{**}}\text{ and }\lim_{n\to\infty}x_n(f)=1,\text{ then }(x_n)_{n=1}^\infty\text{ is norm-convergent in }X^{**}.\end{equation}

Note that we can view any $x_n\in S_X$ as $x_n\in S_{X^{**}}$ with $x_n(f)=f(x_n)$. Moreover, since $X$ is a Banach space, it is complete so that if $(x_n)_{n=1}^\infty$ is norm-convergent in $X^{**}$ then it is also norm-convergent in $X$. Hence we get the following:

Corollary B. Let $X$ be a Banach space. If the norm $\|\cdot\|$ of $X^*$ is Frechet-differentiable at $f\in S_{X^*}$ then the following condition holds: \begin{equation}\tag{3}\text{if }(x_n)_{n=1}^\infty\subset S_X\text{ and }\lim_{n\to\infty}f(x_n)=1,\text{ then }(x_n)_{n=1}^\infty\text{ is norm-convergent in }X.\end{equation}

Now we are ready to prove Theorem 8.6.

Proof of Theorem 8.6. Let $f\in S_X$, which by Corollary 3.56 we just need to show attains its norm. As $\|f\|=1$, we can find $(x_n)_{n=1}^\infty\subset S_X$ such that $f(x_n)\to 1$. By Corollary B, $x_n\to x$ for some $x\in S_X$, and hence $f(x)=1$. $\square$