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Is there a way to prove that the epigraph of any real function $f$ is closed iff $f$ is lower semi-continuous without using limit superior or inferior?

user90335
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    Is your definition of lower semicontinuity that $f^{-1}((c,\infty))$ is open for all $c\in \mathbb{R}$, or the fugly one with $f(x) \leqslant \liminf_{y\to x} f(y)$? – Daniel Fischer Aug 15 '13 at 22:18

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Let $E(f) = \{(x,y) : f(x) \leqslant y \}$ denote the epigraph of $f$. The $E(f)$ is closed if and only if its complement is open.

Suppose $E(f)$ is closed, let $x \in X$ and $y < f(x)$ (so $(x,y) \notin E(f)$). Since $E(f)$ is closed, there is a neighbourhood $U$ of $x$ and an $\varepsilon > 0$ such that $U\times(y-\varepsilon,y+\varepsilon)\cap E(f) =\varnothing$. By the structure of an epigraph, that entails $U\times(-\infty,y+\varepsilon)\cap E(f) = \varnothing$, hence $f(z) \geqslant y+\varepsilon$ for all $z\in U$, and that means $f$ is lower semicontinuous at $x$, since $x$ was arbitrary, $f$ is lower semicontinuous.

Conversely, suppose $f$ is lower semicontinuous. Let$(x,y) \notin E(f)$, i.e. $y < f(x)$. Define $\mu := \frac{y+f(x)}{2}$. By the lower semicontinuity of $f$, $U = \{z \in X : f(z) > \mu\}$ is open. Since$\mu < f(x)$, we have $x \in U$. Then $U\times (-\infty,\mu)$ is an open neighbourhood of $(x,y)$ that doesn't intersect $E(f)$, hence the complement of $E(f)$ is open.

Daniel Fischer
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  • How does $f(x) \geq y + \epsilon$ imply l.s.c. defined as $\forall(x_{0}\in X, \epsilon>0) ~ \exists \delta \text{ s.t.} ||x_{0} - x|| < \delta \implies f(x_{0}) - f(x) < \epsilon$ at the end of the second paragraph?

    How can you deduce from the above definition of l.s.c. that $U$ is open in the last paragraph?

     – Grant Sep 11 '19 at 04:00
  • @Griffy Given $\epsilon > 0$ (and $x_0 \in X$), let $y = f(x_0) - \epsilon$. Then by closedness of $E(f)$ and the structure of epigraphs, there are $\eta > 0$ and $\delta > 0$ such that ${(x,t) : \lVert x - x_0\rVert < \delta, t < y + \eta} \cap E(f) = \varnothing$, which means that for $\lVert x - x_0\rVert < \delta$ we have $f(x) \geqslant y + \eta = f(x_0) - \epsilon + \eta$, and rearranging that inequality yields $f(x_0) - f(x) \leqslant \epsilon-\eta < \epsilon$. For the converse, take $(x_0,y) \notin E(f)$, that is, $y < f(x_0)$. Choose $\epsilon = (f(x_0) - y)/2$. By the definition … – Daniel Fischer Sep 30 '19 at 18:50
  • … in your comment, there is a $\delta > 0$ such that $f(x_0) - f(x) < \epsilon$, or by rearranging and using the definition of $\epsilon$, $f(x) > f(x_0) - \epsilon = y + \epsilon$ holds for all $x$ with $\lVert x - x_0\rVert < \delta$. That says $B_{\delta}(x_0) \times (-\infty, y+\epsilon) \cap E(f) = \varnothing$, and thus the complement of $E(f)$ is a neighbourhood of $(x_0,y)$. Since this holds for every $(x_0,y) \notin E(f)$, the complement of $E(f)$ is open. – Daniel Fischer Sep 30 '19 at 18:50