I want to prove that a sum of subharmonic is subharmonic using the following definition
"Formally, the definition can be stated as follows. Let G be a subset of the Euclidean space ${\mathbb{R}}^n$ and let
$\varphi \colon G \to {\mathbb{R}} \cup \{ - \infty \}$ be an upper semi-continuous function. Then, $\varphi $ is called subharmonic if for any closed ball $\overline{B(x,r)}$ of center x and radius r contained in G and every real-valued continuous function h on $\overline{B(x,r)} $that is harmonic in B(x,r) and satisfies $\varphi(y) \leq h(y)$ for all y on the boundary $\partial B(x,r) $ of B(x,r) we have $\varphi(y) \leq h(y)$ for all y $\in B(x,r)$.
Note that by the above, the function which is identically $-\infty$ is subharmonic"
http://en.wikipedia.org/wiki/Subharmonic_function
As hinted above this definition is more general because we may have $u\notin L^{1}_{loc}$ and so it may not satisfy the MV property eg. log(x) (otherwise the definitions are equivalent using harmonic functions decreasing to u ,which exist from the Dirichlet problem -details omitted-).
Attempts
We start with $u(x)\geq f(x)+g(x)$ for all $x\in \partial B$ for subharmonics f,g.
we want harmonic h s.t. $h(x)\geq g(x)$ and $u(x)-h(x)\geq f(x)$ for all $x\in \partial B$. Then the result follows from the subharmonicity of f,g.
I will type as I think about. One idea is to modify g to get a harmonic function h.
Thanks
Remark: I think the proof by direct definition may not be the easiest.
You may find the following proof in "Counterexamples in Several complex variables" by Fornæss and Stensønes:
First, we start with $f+g\leq u$ on the boundary of $B(x,r)\subset\subset G$, where $f$ and $g$ are subharmonic on $G$, and $u$ is harmonic on some open neighbourhood of $B(x,r)$. Let $\epsilon>0$ be given.
** Then we may find continuous functions $\alpha$ and $\beta$ such that $f\leq \alpha$, $g\leq\beta$ and $\alpha+\beta=u+\epsilon$.
Now, solving the Dirichlet's problem, we may find $F$ and $G$ both harmonic such that $F$ on the boundary of the ball is $\alpha$ and correspondingly for $\beta$. By uniqueness, $F+G=u+\epsilon$.
Since $f\leq\alpha$, by hypothesis that $f$ is subharmonic, $f\leq F$ and likewise $g\leq G$ on $B(z,r)$. Hence $f+g\leq F+G=u+\epsilon$ on $B(z,r)$ and we are done since $\epsilon$ is arbitrary.
The only problem is **, which is not explained but I will try to give a proof. Since $f+g\leq u$, so $f\leq u-g<u-g+\epsilon$. So by compactness of the boundary of the ball, we may probably find $\alpha$ continuous such that $f\leq \alpha\leq u-g+\epsilon$. Hence $g\leq u-\alpha+\epsilon$ and we may choose $\beta=u-\alpha+\epsilon$, which is continuous. I am not even sure if this is correct.