Let $u$ be a continuous function on $\Bbb D$ which is harmonic on $\Bbb D\setminus\{0\}$. If
$$\lim_{|(x,y)|\to 1}u(x,y) = 0,\quad\lim_{|(x,y)|\to 0}u(x,y)/\log(x^2+y^2) = 0,$$
then $u$ is identically zero on $\Bbb D$.

Since $\lim_{|z|\to 0}{u(z)\over\log|z|} = 0$, near $0$, $\epsilon\log|z|\leq u(z)\leq-\epsilon\log|z|$ for some small $\epsilon$. If $|\zeta| =1$ or $\zeta = 0$ then $\liminf_{z\to\zeta}u(z)+\epsilon\log|z|\leq 0$. Hence by the minimum principle of harmonic function, $u(z)\leq -\epsilon\log|z|$ on $\Bbb D\setminus\{0\}$. Similarly, if $|\zeta| =1$ or $\zeta =0$ then $\limin…

@nickbros123 see if this convinces you. Let S be a subset of the reals. If it has a minimum, that's the infimum. Suppose it has no minimum, and suppose it has some lower bound.

Let L be the set of lower bounds of S. Clearly, L is bounded above by the elements of S, so it has a supremum, sup L. Suppose s is an arbitrary element of S.

If s is to the left of sup L, then sup L is not the supremum of L, which is a contradiction. s is not at sup L either, because since S has no minimum, that implies there is another element of S at sup L's left, which we've seen is a contradiction. Therefore, s…