# About the proof of the Ekeland's variational principle

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]?

• I think you are right. $f$ is l.s.c. and $\chi_{x_0}$ is u.s.c.. There is no reason why the sum is l.s.c.. – Kabo Murphy Jul 19 at 23:32
• The problem is the definition on the indicator function. If it is defined as the comment below then $\chi_{x_0}$ is l.s.c. indeed. – The Student Jul 20 at 22:51

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.
• Why the indicator function is l.c.s.? I am agree that $\chi_{x_0}$ is closed. For a counterexample consider $\chi_{x_0} = [ 0 ,1]$ and the sequence defined by $x_n = 1 + \frac{1}{n}$. Then $x_n \longrightarrow x = 1$ and : $$\liminf_{n \to \infty}\chi_{x_0}(x_n) = 0 < 1 = f(x)$$So $\chi_{x_0}$ is not l.s.c – The Student Jul 20 at 15:17
• @Student The indicator function is defined as $I_C(x)=0$ if $x\in C$, $I_C(x)=+\infty$ if $x\not\in C$. It should not be mistaken for a characteristic function, which takes values from $\{0,1\}$. – daw Jul 20 at 20:46