The following is an exercise from Bruckner's Real Analysis book:
A Banach space $X$ has a dual $X^∗$ that is also a Banach space and so has its own dual, denoted by $X^{∗∗}$.
(a) Show that the mapping $φ:X \to X^∗$ defined by $\langle x, x^∗\rangle = \langle x,φ(x) \rangle$ is a linear isometry of $X$ to a closed subspace of $X^{∗∗}$. If $X^{∗∗}= φ(X)$, we say that $X$ is reflexive. (In this case $X$ is isomorphic, in the sense defined in Section 12.10, to its second dual $X^{∗∗}$.)
(b) Prove that $X$ is reflexive if and only if $X^{∗}$ is reflexive.
(c) Prove that if $X$ is reflexive, then every continuous linear functional on $X$ assumes a maximum on the closed unit ball of $X$. [Hint: Use Theorem 12.37 to obtain an element $x^{∗∗}$ of $X^{∗∗}$ such that $\|x^{∗∗}\|=1$ and $\langle x^∗, x^{∗∗} \rangle = \|x^{∗}\|$. Use reflexivity to find $x∈X$ with $\langle x, x^{∗} \rangle = \|x^{∗}\|$].
(d) Refer to Example 12.6. Define $x^{∗}$ as follows: For each element $a = {\{a_1,a_2,...}\}$ of the space $c_0$, we require $\langle a, x^{∗} \rangle = \sum_{k=1}^{\infty} a_k/k!$. Show that $x^{∗} ∈ c_0$ and that $\|x^{∗}\|=\sum_{k=1}^{\infty} 1/k!$. Use part (c) to show that $c_0$ is not reflexive. (In fact it can be shown that $c_0^{∗}=\ell_1$ and $c_0^{∗∗}=\ell_∞$.
(e) Prove that $\mathcal{C}[a, b]$ is not reflexive.
Item (a) : What is $x^*$? $x^*$ is an element of $X^*$ indeed but do we choose it arbitrarily or is determined by $x$ in which case there are different theorem gives different $x^*$ based on one fixed $x$.
Item (b) : $\phi(X^*)$ is not defined to define reflexivity of $X^*$ so how to begin proving when it is not defined clearly?
Item (c) : it is referred to Theorem 12.37 which is lacks inner product! How $ \langle x^*,x^{**} \rangle$ appears from Theorem 12.37 so that to initiate solving the problem?
Similar problems with other items.
Item a: There are two typos in the statement of item a of this exercise. The mapping is $\phi : X \rightarrow X^{**}$ and it is defined as: for all $x \in X$, $\phi(x)$ is a functional defined on $X^*$ as, for all $x^* \in X^*$, $$ \langle x, x^*\rangle = \langle x^*, \phi(x)\rangle $$
that is $\phi(x)(x^*) = x^*(x)$, for all $x^* \in X^*$.
Let us prove that $\phi$ is a linear isometry of $X$ into a closed subspace of $X^{**}$.
Proof:
Step 1. Given any $x \in X$, It is easy to seem that $\phi(x)$ is a linear functional defined on $X^*$. Moreover, for all $x^* \in X^*$, $$|\phi(x)(x^*)| = |x^*(x)| \leq \|x^*\|\|x\|= \|x\|\|x^*\| \tag{1}$$ So, $\phi(x)$ is a bounded linear functional, so $\phi(x) \in X^{**}$.
So, we have that $\phi$ is actually a mapping from $X$ to $X^{**}$.
Step 2. It is easy to check that $\phi$ is linear. In fact, given $\alpha$ scalar and $x , y \in X$, we have, for all $x^* \in X^*$, \begin{align*} \phi(\alpha x +y)(x^*) &= x^*(\alpha x +y) = \alpha x^*(x) + x^*(y) = \\ & =\alpha \phi(x)(x^*) +\phi(y)(x^*) = (\alpha \phi(x) +\phi(y))(x^*) \end{align*} So $\phi(\alpha x +y) = \alpha \phi(x) +\phi(y)$. So $\phi$ is linear.
Step 3: Given any $x \in X$, using $(1)$, we have that, $\|\phi(x)\| \leq \|x\|$ (and so $\phi$ is a continuous linear mapping). Using Theorem 12.37, we know that there is $x^*\in X^*$ such that $\|x^*\|=1$ and $$ \phi(x)(x^*) = x^*(x) = \|x\|= \|x\|\|x^*\|$$ So $\|\phi(x)\| = \|x\|$. So, $\phi$ is a linear isometric mapping from $X$ to $X^{**}$.
Step 4 The image of $\phi$ is a closed subspace of $X^{**}$. In fact, let $x^{**} \in X^{**}$ and let $\{x^{**}_n\}_n$ be a sequence in the image of $\phi$ converging to $x^{**}$. Then, $\{x^{**}_n\}_n$ is a Cauchy sequence.
Let $\{x_n\}_n$ in $X$ be a sequence such that, for all $n$, $\phi(x_n) = x^{**}_n$. Since $\phi$ is an isometric transformation, we have that $\{x_n\}_n$ is a Cauchy sequence in $X$. So, $\{x_n\}_n$ converges to some $x \in X$. Since $\phi$ is continuous, $$\phi(x) = \phi(\lim_n x_n) = \lim_n(\phi(x_n)) = \lim_n x^{**}_n = x^{**}$$ So, $x^{**}$ is in the image of $\phi$. So, the image of $\phi$ is a closed subspace of $X^{**}$.
Item b:
$(\Rightarrow)$ If $X$ is reflexive, then $X$ is isomorphic (as a Banach space) to $X^{**}$. It follows that $X^*$ is isomorphic (as a Banach space) to $(X^{**})^*$. Now, note that $ (X^{**})^{*} = ((X^*)^*)^*= (X^{*})^{**}$. So, $X^*$ is isomorphic (as a Banach space) to $(X^{*})^{**}$. So, $X^*$ is reflexive.
$(\Leftarrow)$ Suppose that $X$ is not reflexive. Let $\phi : X \rightarrow X^{**}$ be the isometric linear injection of item a. Then $I = \phi(X)$ is a proper closed subspace of $X^{**}$. By a corollary of Hahn-Banach (see Remark), there is $\ell \in X^{***}$ such that $\ell \ne 0$ and $\ell|_I=0$.
Now, if $X^*$ is reflexive, then there is $x^* \in X^*$ such that, for all $x^{**} \in X^{**} $, $\ell(x^{**})= x^{**}(x^*)$. Since $\ell \ne 0$, we have thet $x^* \ne 0$. However, since $\ell|_I=0$, we have that, for all $x \in X$, $$0 = \ell(\phi(x))= (\phi(x))(x^*) = x^*(x) $$ So $x^* =0$. Contradiction. So, $X^*$ is not reflexive.
Item c: In Banach space, the notation $\langle x, x^* \rangle$ does not mean inner product. It is just a notation for $x^*(x)$. See comments on page 504, just before Theorem 12.35. Once you know it, the hint for this item c is easy to follow. In fact, this item is a direct consequence of Theorem 12.37 and the reflexivity.
Item d: Condider $c_0$ (with the with the supremum norm). There is one typo in the statement of this item. we must prove that $x^* \in c_0^*$. Note that for all $a=\{a_1, a_2, \dots \} \in c_0$, we have $$ |x^*(a)| = |\langle a, x^* \rangle| = \left | \sum_{k=1}^\infty a_k / k! \right |\leq \left ( \sum_{k=1}^\infty 1/ k!\right)\|a\|_\infty$$
So, $x^* \in c_0^*$ and $\|x^*\| \leq \sum_{k=1}^\infty 1/ k!$. Let $s_n$ be a sequence such that the $n$ first position are $1$ and that other positions are $0$. Then $\|s_n\|_\infty = 1$, and, for all $n$,
$$ |x^*(s_n)| = \sum_{k=1}^n 1 / k! $$ So, $\|x^*\| = \sum_{k=1}^\infty 1/ k!$.
Finally, note that there is no $x$ such that $\|x\|_\infty\leq 1$ (that is, $x$ in the closed unit ball of $c_0$) and $$ x^*(x) = \sum_{k=1}^\infty 1 / k! $$ So, using item c, we see that $c_0$ is not reflexive.
Item e: Given $a < b$, consider $X=\mathcal{C}[a,b]$ (with the with the supremum norm). Let $d=(a+b)/2$ and let $x^*$ be defined as $$ x^*(f) = \int_a^b f \cdot(\mathbf{1}_{[a,d]} - \mathbf{1}_{[d,b]}) \ d\lambda$$ where $\lambda$ is the Lebesgue measure. It is easy to see that $x^*$ is a continuous linear functional. It is also easy to see that $\|x^*\| = b-a$. However, there is no $f \in \mathcal{C}[a,b]$ such that $\|f\|_\infty \leq 1$ and $ x^*(f)=b-a$. So, by item c, $\mathcal{C}[a,b]$ is not reflexive.
Remark: For a direct proof of this step: If $I$ is a closed subspace of $X^{**}$, take a vector in $v$ in $X^{**}$ but not in $I$. Define $\ell_0$ manually to be zero on $I$ and such that $\ell_0(kv) = k\|v\|$. It is easy to see that $\ell_0$ is a non-zero bounded linear functional defined on the subspace generated by $I$ and $v$. Use Hahn-Banach to extend $\ell_0$ to the whole space $X^{**}$. Such extension is $\ell$ .