I am putting together a proof for the Harnack inequality for harmonic functions defined on a balls. I would like to do this using the mean value property of Harmonic functions.
I am thinking of doing this. Let us fix $x_0$, $y_0$ in some ball of radius $r$ centered around the origin. Now, we will construct two sub-balls around each of these balls. Specifically, let us take some arbitrarily small $\epsilon$ ball around $x_0$. Now, if we take $\epsilon$ small enough we can contain this sub-ball in a ball $B_{r_{y_0}} (y_0)$ around $y_0$ such that the ball around $y_0$ is a subset of our original ball $B_r$ on which the function is defined/harmonic. The mean value property then tells us that: $$ C(\epsilon) \cdot u (x_0) = \int_{B_{\epsilon}(x_0)} u(x) \mathrm{d}x $$ and moreover: $$ C(r_y) \cdot u(y_0) = \int_{B_{r_{y}}(y_0)} u(y) \mathrm{d}y $$ Moreover, because the domain of the $y$ integral contains the domain of the $x$ integral: $$ C(r_y) \cdot u(y_0) = \int_{B_{r_{y}}(y_0)} u(y) \mathrm{d}y \geq \int_{B_{\epsilon}(x_0)} u(x) \mathrm{d}x = C(\epsilon) \cdot u (x_0) $$ Hence: $$ \frac{C(r_y)}{C(\epsilon)} u(y_0) \geq u(x_0) $$ Now, my only issue with the above proof is dependencies. Specifically, it seems like my choice of constants determined is a function of the initial points I choose, which does not seem to be correct (through the general statement of the inequality). Could someone please help me verify if this is correct?
This is inevitable. The constant in Harnack's inequality depends on the points involved. Usually it is stated as "for each compact subset $K$ of the domain there is a constant $C_K$ such that $u(x)/u(y)\le K$ for every $x, y\in K$ (where $u$ is a positive harmonic function)" but this doesn't really change anything: one has to take some $K$ that contains the given $x, y$.
More importantly, your proof breaks down at this step:
This cannot be done in general. For example, $x_0$ and $y_0$ could be these red dots: there is no ball centered at $y_0$ that is contained in the original ball $B$ and contains $x_0$.
However, the argument can be repaired. It is valid when $|x_0-y_0| \le \frac12 \operatorname{dist}(y_0, \partial B)$, because one has $B(x_0, r)\subset B(y_0, 2r)\subset B$ where $r = |x_0-y_0|$.
For general points, connect $x_0$ to $y_0$ by a line segment (in a general domain, use some curve) and place a finite sequence of points on that segment, locating them close enough to each other so that above applies.