Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 2 / d

Assume condition (1). For fixed $a$, the supremumis nondecreasing as $r$ increases. Therefore its limit as $r\downarrow 0$ exists in the extended real numbers and equals ${\inf }_{r>0}{M}_a(r)$. Because $a\in B(a,r)$, it is at least $f(a)$. Condition (1) says precisely that this limit is $f(a)$.

It follows that for every $\epsilon >0$ there is $r>0$ with $M_a(r)<f(a)+\epsilon$. Fix $C\in \mathbb{R}$ and $a$ with $f(a)<C$. Taking $\epsilon =C-f(a)$ produces an open ball on which $f(x)<C$. Thus $\{x:f(x)<C\}$ is an open set, and its complementThis proves (1) implies (2). The property being characterized is upper semicontinuity.

Assume condition (2). Fix $a\in X$ and $\epsilon >0$. The superlevel setis a closed set and does not contain $a$. Its complement is therefore an open neighbourhood of $a$, so some open ball $B(a,r)$ avoids $F$. Consequently $f(x)<f(a)+\epsilon$ throughout that ball, givingThe same bound holds for every smaller radius. Taking $r\downarrow 0$ and then $\epsilon \downarrow 0$ yieldsTogether with the previous implication this proves the equivalence. A supremum may equal $f(a)+\epsilon$ even when every individual value is strictly smaller; the non-strict estimate here handles that possibility.

If $f$ is continuous at $a$, then for every $\epsilon >0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\epsilon$. In particular, for $0<r\le r_0$,The squeeze theorem giveswhich implies condition (1). Thus continuity implies upper semicontinuity.

Apply the local bound from condition (1) separately to $f$ and $-f$. For any $a\in X$ and $\epsilon >0$, there are $r_{+},r_{-}>0$ such thatWith $\delta =\min \{r_{+},r_{-}\}$, both inequalities hold when $d(x,a)<\delta$, soThis is exactly continuity at $a$. Since $a$ was arbitrary, $f$ is continuous on $X$: upper semicontinuity of $f$ and of $-f$ is equivalent to simultaneous upper semicontinuity and lower semicontinuity of $f$.