About the proof of the Ekeland's variational principle

Question

Ekeland's variational principle statement: Let $(X, \| \cdot \|)$ be a Banach space and $f: X \longrightarrow \mathbb{R} \cup \{ +\infty \}$ an extended-valued proper lower semicontinuous function, bounded from below. Then for each $\epsilon > 0$ there exists $x^* \in X$ such that: \begin{equation*} f(x^*) < f(x) + \epsilon \| x^* - x \| \; \forall x \in X \setminus \{x^*\} \qquad...[1] \end{equation*} And for a given $x_0 \in X$, the former $x^* \in X$ can be chosen to meet: \begin{equation*} f(x^*) + \epsilon \| x^* - x_0 \| \leq f(x_0) \qquad...[2] \end{equation*} According to an article, for the proof of this theorem once [1] is proved, it can be assumed in to obtain [2] with $\hat{f} := f + \chi_{x_0}$, with $\chi_{x_0}$ the indicator function on the set \begin{equation*} X_0:= \{ z \in X: f(z) + \epsilon \| z - x_0 \| \leq f(x_0) \} \end{equation*} I have the strong belief that $\hat{f}$ is not necessarily lower semicontinuous. In the opposite case for applied $[1]$ why $x^* \in \chi_{x_0}$? If the recommendation is useless how you can use [1] to prove [2]?

Answer 1

The function $z\mapsto f(z) + \epsilon \|z-z_0\|$ is l.s.c, so its lower level sets are closed, and the associated indicator function is l.s.c. as well. Sums of l.s.c. functions are l.s.c., so $\hat f$ is l.s.c.