I'm reading part 1.4 in textbook Functional Analysis, Sobolev Spaces and Partial Differential Equations. The author presents these equivalents definitions without proof. Could you verify if my proof is fine?
Let $X$ be a topological space and $f:X \to \mathbb R \cup \{-\infty, +\infty\}$. For $x \in X$, let $\mathcal N(x)$ be the set of all open neighborhoods of $x$. We define $$\liminf _{y \to x} f(y) := \sup _{V \in \mathcal N (x)} \inf _{y \in V \setminus \{x\}} f(y) , \quad x \in X.$$
Then the following statements are equivalent.
(a) For all $\lambda \in \mathbb R$, the set $\{ y \in X \mid f(y) \le \lambda\}$ is closed in $X$.
(b) For every $(x, \lambda) \in X \times \mathbb R_{>0}$, there is a neighborhood $V$ of $x$ such that $f(y) \ge f(x) - \lambda$ for all $y \in V$.
(c) The epigraph of $f$, defined by $\operatorname{epi} f := \{ (x, \lambda) \in X \times \mathbb R \mid f(x) \le \lambda\}$, is closed in the product topology of $X \times \mathbb R$.
(d) For all $x \in X$, $$f(x) \le \liminf _{y \to x} f (y).$$
My attempt:
- (a) implies (b)
Let $(x, \lambda) \in X \times \mathbb R_{>0}$. The set $U:=\{y \in X \mid f(y) > f(x) - \lambda\}$ is open and $x \in U$, so there is a neighborhood $V$ of $x$ such that $V \subseteq U$, i.e., $f(y) > f(x) - \lambda$ for all $y \in V$.
- (b) implies (c)
Let $(x, \lambda) \in (\operatorname{epi} f)^c$. We choose $\lambda' > 0$ such that $f(x) - \lambda' >\ \lambda$. Such $\lambda'$ exists because $f(x) > \lambda$. There is a neighborhood $V$ of $x$ such that $f(y) \ge f(x) - \lambda'$ for all $y \in V$. Let $\epsilon := (f(x) - \lambda' - \lambda)/2 > 0$. Then $f(y) > \lambda''$ for all $y \in V$ and $\lambda'' \in U := ( \lambda - \epsilon, \lambda + \epsilon)$. It follows that $V \times U$ is a neighborhood of $(x, \lambda)$ such that $V \times U \subseteq (\operatorname{epi} f)^c$. Hence $(\operatorname{epi} f)^c$ is open.
- (c) implies (d)
Given $\varepsilon > 0$, let $\lambda := f(x) - \varepsilon$. Then $(x, \lambda) \in (\operatorname{epi} f)^c$. Then there is an open neighborhood $V$ of $(x, \lambda)$ such that $f(y) > \gamma$ for all $(y, \gamma) \in V$. Let $I_\varepsilon := (\lambda - \varepsilon, \lambda + \varepsilon)$. Then $V_\varepsilon := (X \times I_\varepsilon ) \bigcap V$ is also open. Let $V_x$ and $V_\lambda$ be the images of $V_\varepsilon$ through the projection maps. Then $V_x$ and $V_\lambda$ are also open. We have $$f(y) > \inf V_\lambda \ge \lambda - \varepsilon = f(x) - 2\varepsilon, \quad \forall y \in V_x.$$
It follows that $$f(x) - 2\varepsilon \le \inf _{y \in V_x \setminus \{x\}} f(y) \le \liminf _{y \to x} f (y).$$
Take the limit $\varepsilon \to 0$, we obtain $$f(x) \le \liminf _{y \to x} f (y).$$
- (d) implies (a)
Let $\lambda \in \mathbb R$ and $x \in X$ such that $f(x) > \lambda$. This implies $\lambda < \liminf _{y \to x} f (y)$. Hence there is an open neighborhood $V$ of $x$ such that $$\lambda < \inf _{y \in V \setminus \{x\}} f(y).$$
This in turn implies $f(y) > \lambda$ for all $y \in V$. Hence $\{ y \in X \mid f(y) > \lambda\}$ is open and thus $\{ y \in X \mid f(y) \le \lambda\}$ is closed.
