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I am not able to finalize the proof of this statement, and it seems something is always omitted in textbooks. Here there is what I got so far. (I'm trying to find the proof for a classical real-valued function, not for extended real-valued functions, I hope this is not an obstacle).

A function $f: \text{dom}f \subseteq X \to \mathbb{R} $ ( $X$ is a topological space) is said to be closed if its epigraph is closed: $$\text{epi}f = \{(x,y) \in \text{dom}f \times \mathbb{R}: y \ge f(x)\} = \{(x,y) \in \text{dom}f \times \mathbb{R}: f(x) \le y \} = \{(x,y) \in \text{dom}f \times \mathbb{R}: x \in S_yf\}$$ where $ S_cf = \{x \in \text{dom}f : f(x) \le c\}$ is the $c$ sublevel set of $f$.

For a certain $y\in\mathbb{R}$, then $(x,y) \in \text{epi}f \Leftrightarrow x \in S_yf $ so $\text{epi}f$ is closed iff $S_yf $ is closed.

Now, everywhere it is said that a necessary and sufficient condition for a function to be closed, is that lower semi-continuity holds for each points in the domain:

$$\forall x_0 \in \text{dom}f, \forall \epsilon >0, \exists \delta > 0: f(x)\ge f(x_0) - \epsilon, \forall x \in B_\delta(x_0)$$ where $B_\delta(x_0)$ is the open ball centered in $x_0$ with radius $\delta$.

I'm already stuck with the sufficiency part.

Let's consider a point $\bar x \in \partial S_yf $, then from the definition of boundary of $S_yf$: $$\exists R>0: f(x) \le y, \forall x \in S_yf \cap B_R(\bar x) \setminus \{\bar x\}$$ where $S_yf \cap B_R(\bar x) \setminus \{\bar x\}$ is non empty.

If $\bar x \in \text{dom}f$, using the lower semicontinuity of $f$ in $\bar x$: $$\forall \epsilon >0, \exists \delta > 0: f(\bar x)\le f(x) + \epsilon, \forall x \in B_\delta(\bar x)$$

Since $f(x) \le y, \forall x \in B_R(\bar x) \cap S_yf \Longrightarrow f(x) + \epsilon \le y + \epsilon , \forall x \in S_yf \cap B_\delta(\bar x) \cap B_R(\bar x) \setminus \{\bar x\} $

where $ S_yf \cap B_\delta(\bar x) \cap B_R(\bar x) \setminus \{\bar x\} $ is again non empty.

Then $f(\bar x)\le f(x) + \epsilon \le c + \epsilon > 0, \forall \epsilon$ therefore, necessarily $f(\bar x)\le c \Rightarrow \bar x \in S_yf$.

If instead the point does not belong to the domain: $\bar x \notin \text{dom}f$, and, of course, $\bar x \notin S_yf $, so one must prove that this point cannot exist. The only thing I can prove is that, the function cannot have limit $+\infty$ in $\bar x$.

Again, $\exists R>0: f(x) \le y, \forall x \in S_yf \cap B_R(\bar x) \setminus \{\bar x\}$.

Let us consider any sequence $x_k \in \text{dom}f$ such that $x_k \in S_yf \cap B_R(\bar x) \setminus \{\bar x\}, \forall k \ge k_0$ such that: $$\lim_{k\to \infty} x_k = \bar x$$ then, since $f(x_k)\le y, \forall k \ge k_0$, this limit, if existing, can be $y$ at most.

Apparently I'm struggling to prove something that is not even true. As these slides(pages 5-7) from a MIT course state, the equivalence holds for functions having all $\mathbb{R}^n$ as their domain (or, in general, all $X$ topological space) and also:

"If $f$ is lower semicontinuous at all $x \in \text{dom}f$, it is not necessarily closed".

I am stil a bit confused by the last paragraph in "Formal definition" of semicontinuity Wikipedia page

EDIT:

I said "from definition of boundary", meaning simply I can always find some points in any neighborhood of a boundary point that belong to the set, so I used the intersection between a ball around a boundary point and the set itself, claiming that such a set is non empty (thanks to the definition of boundary). Only now I figure out that this statement is not so clear to read.

As regards the structure of the set $X = \mathrm{dom} f$, I don't understand which properties have to be satisfied in order to consider it a topological space (maybe my topology background is not so thick, I'm sorry, I'm just attending a Convex analysis and optimization master course). The proof of the first implication is clear enough for me, whereas in the second one ($2 \Rightarrow 1$) I don't get why the set $\{(x,y)\colon x\in X\}\subset X\times\Bbb R$ should be closed for each $y$: if $X$ is open how can this set be closed. Moreover the set $\{(x,y)\colon x\in X,f(x)\le y\}$ seems to me the epigraph itself. Regarding the next statement about the projection of this set over the set $X$ I think I'm missing the property that lets us infer the closedness after the projection.

As you pointed out, I guess the key point lies in the statement "$f$ is closed in $X$", that I think it is supposed to mean that even if $X$ is open it doesn't matter because the property must hold in any closed set inside it, but I am not sure to get this point. Even if I have already enough problems in properly understanding your explaination (sorry about that), I would like to make a very naive example which maybe could clarify better my not so clever perplexity. Let us consider the function $f\colon B_{R}(0) \subset \Bbb R^{2} \to \Bbb R, f(x) = \| x \| ^2$. This function is lower semicontinuous (actually even continuous over the open ball $B_{R}(0)$) but it is not closed, because for instance any sublevel set $S_c$ with $c \ge R^2$ is an open set.

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  • $\begingroup$ To resolve the confusion: for $f\colon X\to [-\infty,+\infty]$ to be semicontinuous is equivalent to epigraph being closed $\color{red}{\text{in }X\times\Bbb R}$. In this sense, $f\colon (0,1)\to\Bbb R$ defined as identical zero has the closed epigraph as it is considered in the topology of $(0,1)\times\Bbb R$. On the MIT slides they consider it in the topology of $\Bbb R^2$, which is a bit strange. $\endgroup$ – A.Γ. May 12 '17 at 16:04
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This definition of the boundary of $S_yf$ that you use $$ \exists R>0: f(x) \le y, \forall x \in S_yf \cap B_R(\bar x) \setminus \{\bar x\} $$ looks strange to me. It says that locally near $\bar x$ all $x$ from $S_yf$ should satisfy $f(x)\le y$, but this is true for all $x\in S_yf$ globally (by definition of the sublevel set), and it has nothing to do with the boundary.

Since you prefer to work with the finite values of $f$ we take $X=\operatorname{dom}f$. If $\operatorname{dom}f$ is a subset of a larger topological space then it becomes a topological space itself when equipped with the induced topology. Assuming $X$ to be Hausdorff we prove that the following statements are equivalent:

  1. $f\colon X\to\Bbb R$ is lower semicontinuous,
  2. $\operatorname{epi}f$ is closed in $X\times\Bbb R$.

Proof:

$\fbox{$1\Rightarrow 2$}$

Let $(x_\alpha,y_\alpha)_{\alpha\in A}$ be a net in $\operatorname{epi}f$ that converges to $(\bar x,\bar y)\in X\times\Bbb R$. We need to prove that $(\bar x,\bar y)\in\operatorname{epi}f$. But it is easily true because $$ f(\bar x)\le\liminf f(x_\alpha)\le\liminf y_\alpha=\bar y. $$ The first inequality is due to $f$ being lower semicontinuous, the second one comes from $(x_\alpha,y_\alpha)\in\operatorname{epi}f$.

$\fbox{$2\Rightarrow 1$}$

The set $\{(x,y)\colon x\in X\}\subset X\times\Bbb R$ is closed for each $y\in\Bbb R$. Hence, since $\operatorname{epi}f$ is closed, the intersection $$ \operatorname{epi}f\cap\{(x,y)\colon x\in X\}=\{(x,y)\colon x\in X,f(x)\le y\} $$ is closed too for each $y\in\Bbb R$. It follows that for each $y\in\Bbb R$ the projection on $X$ $$ S_y=\{x\colon x\in X,f(x)\le y\} $$ is closed.

Now take a net $(x_\alpha)_{\alpha\in A}$ in $X$ that converges to $\bar x\in X$ and denote $\bar y=\liminf f(x_\alpha)$. This limit cannot be $-\infty$, because it would mean that $\bar x\in\operatorname{closure}(S_y)=S_y$ for every $y\in\Bbb R$, which is impossible (as $f(\bar x)\in\Bbb R$). Hence, $\bar y>-\infty$ and $\bar x\in\operatorname{closure}(S_{\bar y+\epsilon})=S_{\bar y+\epsilon}$ for all $\epsilon>0$. Taking $\epsilon\to 0$ gives that $f(\bar x)\le \bar y$. Hence, the function is lower semicontinuous.

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  • $\begingroup$ First of all, I want to thank you for your patience and helpfulness. See my revised question. $\endgroup$ – Vexx23 May 14 '17 at 14:31
  • $\begingroup$ @Vexx23 Most of your questions concerns with the same thing: how $X=(0,1)$ can be closed if $(0,1)$ is open? Yes, $X$ is open in $\Bbb R$, but here we consider the subspace topology (see link examples). Imagine that there is no $\Bbb R$, and $(0,1)$ is the whole world that you have. Then $(0,1)$ is closed as the whole space must be closed in any topology (and open as well) since the empty set is open. Hence, $X\times \{y\}$ is closed, $\{(x,y)\colon x\in X,f(x)\le y\}$ is closed (it is not the epigraph, it is a part of it for the fixed $y$). $\endgroup$ – A.Γ. May 15 '17 at 0:11
  • $\begingroup$ @Vexx23 In your example, the sublevel set for $c\ge R^2$ is again the whole space ($=B_R(0)$). It is both closed and open $\color{red}{\text{in }B_R(0)}$, because there is nothing outside it, the function is undefined outside the ball. $\endgroup$ – A.Γ. May 15 '17 at 0:13
  • $\begingroup$ Now it is more clear, I am not familiar with the concept of subspace topology. In any case, in the above theorem the function must have the whole topological (sub) space as its domain, I get this is the key point. $\endgroup$ – Vexx23 May 16 '17 at 16:04

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