Problem in understanding maximum principle for subharmonic functions

Question

I am reading subharmonic functions and their properties from the book From Holomorphic Functions to Complex Manifolds by Grauert and Fritzsche. Let me first define what a subharmonic function is.

According to the book I mentioned, a function $s : G \longrightarrow \mathbb R \cup \{-\infty\}$ (where $G \subseteq \mathbb C$ is a domain) is said to be subharmonic if the following conditions hold $:$

$(1)$ $s$ is upper semicontinuous.

$(2)$ For every disk $D \subset \subset G$ (i.e. $\overline D \subseteq G$ is compact) if there exists a continuous function $h : \overline D \longrightarrow \mathbb R$ with $h \lvert_{D}$ harmonic and $s \leq h$ on $\partial D,$ then $s \leq h$ on $D.$ This property is known as the harmonic majorant property.

An interesting fact about subharmonic functions is that they also satisfy maximum principle as in the case of harmonic functions. The proof of this given in the book is the following $:$

The proof is quite clear and understandable modulo the statement highlighted in yellow. I have searched many properties of upper semicontinuous functions in google but I didn't able to find the property highlighted in yellow. Could anyone have any insight on how does the claim work? Any suggestion would be warmly appreciated.

Thanks for your time.

Answer 1

Proposition $:$ If $f$ is a u.s.c. function on $\Omega$ and bounded above, then there is a sequence $f_1 \geq f_2 \geq \cdots$ of continuous functions on $\Omega$ that are bounded above and that converge to $f.$

For a proof of the above proposition please have a look at Proposition 2.1.2 in chapter $2$ from the book Function Theory of Several Complex Variables by Steven G. Krantz.

In your case since $\partial D$ is compact and $s$ is u.s.c. it follows that $s$ is bounded above. So by the above proposition there exists a decreasing sequence of continuous functions $\{f_j\}_{j \geq 1}$ on $\partial D$ such that $f_j \downarrow s$ pointwise on $\partial D$ and hence $s \leq f_j$ for all $j \geq 1.$ Now assume that $s(b) \in \mathbb R$ and let $r : = c - s(b)$ and choose $0 \lt \varepsilon \lt r.$ Then there exists $j_0 \geq 1$ such that $f_{j_0} (b) \lt s(b) + \varepsilon \lt s(b) + r = c.$ So $h = f_{j_0}$ will do the job for us. Otherwise if $s(b) = - \infty$ then $f_j(b) \downarrow - \infty$ and hence there exists $j_1 \geq 1$ such that $f_{j_1} (b) \lt c.$ So in this case we may take $h = f_{j_1}.$

Answer 2

While this particular case is easier (as shown by Arnab), it also follows from the very useful, and not all that well known Katětov–Tong insertion theorem stating that:

If $X$ is a normal topological space and $g \le h$ are real-valued functions on $X$ where $g$ is upper semicontinuous and $h$ is lower semicontinuous, then there exists a continuous $f\colon X \to \mathbb{R}$ with $g \le f \le h$.