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The strong maximum principle for subharmonic functions states that an upper semicontinuous subharmonic function on a connected open set which attains its finite global maximum at an interior point is constant. In particular this holds for a harmonic function; negating the function gives the corresponding minimum principle for superharmonic functions.

Let the maximum be $M$, attained at $x_0$. For every sufficiently small ball centered at $x_0$, the mean value inequality givesTherefore its average equals $M$. If some point in the ball had value less than $M$, upper semicontinuity would give an open neighborhood where $u\le M-\eta$ for some $\eta >0$. That positive-measure neighborhood would make the average strictly less than $M$, a contradiction. Thus $u=M$ throughout a neighborhood of $x_0$.

The same argument at each maximizing point shows that $\{u=M\}$ is open; it is also closed relative to the domain by upper semicontinuity. It is nonempty, so connectedness forces it to be the whole domain. Without connectedness, constancy is only guaranteed on the component containing the maximizing point.

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.

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 $f$.

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.