Removable singularity for harmonic functions

Solution 1:

(1) Let $B(x_0, r) = \{x: |x-x_0|< r\}$. Fix $r$ such that $\overline{B(x_0, r)}\subset \Omega$. Let $v$ be the harmonic function in the $B(x_0, r)$ that agrees with $u$ on $\partial B(x_0, r)$, such a function exists as it's given by the Poisson integral. For $\epsilon>0$, consider $$ w(x) = u(x)-v(x)-\epsilon \Phi(x, x_0) $$ which is harmonic in $B(x_0, r)\setminus \{x_0\}$ and tends to $-\infty$ as $x\to x_0$. By the maximum principle (you may want to remove a tiny neighborhood of $x_0$ before applying it), $$ \sup \{w(x): x\in B(x_0, r)\setminus \{x_0\}\}) = -\epsilon \sup_{|x-x_0|}\Phi(x, x_0) $$ Letting $\epsilon\to 0$, we obtain that $u(x)-v(x) \le 0$ for every $x\in B(x_0, r)\setminus \{x_0\}$ (think of such $x$ as fixed while $\epsilon\to 0$).

The same argument applies to $w(x) = v(x)-u(x)-\epsilon \Phi(x, x_0)$ and shows that $v(x)-u(x) \le 0$ for every $x\in B(x_0, r)\setminus \{x_0\}$. Conclusion: $v$ provides the required harmonic extension of $u$.

(2) There is a benefit in having weaker assumptions in a theorem, because a weaker statement may be easier to verify. You noted that replacing "bounded" by "$o(\Phi)$" does not expand the set of functions to which the theorem applies; but it can make it easier to prove that a particular function falls in this set. The various forms of Phragmén–Lindelöf principle are another example of this. E.g., if a function $f$ is holomorphic in the the strip $|\operatorname{Im} z|<\pi/2$, its boundary values are bounded by $1$, and $$f(z)< \exp(A\exp(c|\operatorname{Re} z|))$$ for some finite $A$ and $c<1$, then $f$ is actually bounded by $1$ in the strip. Why such a fussy condition, when the function is in fact bounded? It's so that we don't have to prove that it's bounded in order to apply the theorem.