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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]?

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  • $\begingroup$ 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.. $\endgroup$ – Kabo Murphy Jul 19 at 23:32
  • $\begingroup$ 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. $\endgroup$ – The Student Jul 20 at 22:51
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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.

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  • $\begingroup$ 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 : \begin{equation} \liminf_{n \to \infty}\chi_{x_0}(x_n) = 0 < 1 = f(x) \end{equation}So $\chi_{x_0}$ is not l.s.c $\endgroup$ – The Student Jul 20 at 15:17
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    $\begingroup$ @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 $\{0,1\}$. $\endgroup$ – daw Jul 20 at 20:46
  • $\begingroup$ Thanks, with this definition I could prove the implication. I will post it later. $\endgroup$ – The Student Jul 20 at 22:58

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