I have a trouble about an exercise in Techniques of Variational Analysis, Borwein, J.M., Zhu, Q.J (Ex. 2.1.2): Let $X$ be a Banach space and let $f: X \to \mathbb{R}$ be a Fréchet differentiable function (https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivativeFréchet_derivative). Suppose that $f$ is bounded from below on any bounded set and satisfies $$\lim_{\left \| x \right \| \to \infty} \frac{f\left (x \right )}{\left \| x \right \|}=+\infty$$ Then the range of $f'$ is dense in $X^*$.
By the similar method in $\mathbb{R}$, for $\gamma \in X^*$, letting $g\left ( x \right ) = f \left ( x \right ) -\langle \gamma , x \rangle $. I proved $g \to \infty$ as $x \to \infty$. However, I have no idea to continue or construct a sequence in the range of $f'$ converging to $\gamma$. I also don't know how to use the hypothesis that $f$ is bounded from below on any bounded set.
You can get the conclusion by applying Ekeland's variational principle to $g$. (The principle ought to be discussed in that book, given its title.)
Since $g(x) = f(x)-\langle \gamma,x\rangle$ is bounded below, for any $\epsilon>0$ you can pick $u\in X$ such that $g(u)<\inf g+\epsilon$. From Ekeland's principle you get the existence of a point $v$ such that $$g(w)-g(v)\ge -\epsilon \|w-v\|\quad \text{ for all }w\tag{1}$$ Suppose $\|g'(v)\|>\epsilon$. Then there exists a unit vector $h$ such that $\langle g'(v), h \rangle > \epsilon$. Hence $$ g(v-th)-g(v) = -t\langle g'(v), h \rangle + o(t) \tag{2} $$ which is strictly less than $ -\epsilon t$ when $t$ is small enough. This contradicts (1), proving that $\|g'(v)\| \le\epsilon$.
Thus $0$ is in the norm-closure of the range of $g'$, hence $\gamma$ is in the norm-closure of the range of $f'$.