Hartogs lemma for subharmonic functions

Remarks

Proof idea: fix $x_0 \in K$ and a small radius $r$ such that $f < f(x_0) + \varepsilon/3$ on a closed ball about $x_0$ inside $\Omega$. By the sub-mean value property, $u_k(y)$ is at most the mean of $u_k$ over the ball $B(y, r)$, for $y$ near $x_0$. The uniform upper bound allows the reverse form of Fatou's lemma, which bounds the limit superior of these ball means by the mean of $\limsup_k u_k \le f$. This gives $u_k \le f(x_0) + \varepsilon$ on a neighbourhood of $x_0$ for large $k$, and compactness of $K$ finishes the proof.

The lemma upgrades a pointwise upper bound to a uniform one on compact sets, in the upward direction only. A typical use: if $u_k \le 0$ in the limit at every point, then $\max_K u_k$ is eventually at most $\varepsilon$ on each compact $K$. The lower bound is not upgraded in the same way, because subharmonic functions can take very negative values on small sets.

Going deeper

For a student

The lemma is named after Hartogs, who used an argument of this kind in his theorem on separately holomorphic functions in several complex variables. It is a basic compactness tool for subharmonic and plurisubharmonic functions. Typical uses are bounds on $\frac{1}{k}\log\abs{p_k}$ for sequences of polynomials $p_k$, and the study of upper envelopes of families of subharmonic functions.