tangent plane to graph of a harmonic function Suppose $u \in C^2(\omega)$ harmonic, with $\omega$ open and connected. Let $\{(x,y,v(x,y)), (x,y) \in \omega\} $ be the tangent plane in $(x_0,y_0)\in \omega$ to the graph of $u$,  $G(u)=\{(x,y,u(x,y)), (x,y) \in \omega\} $. One has to prove that the tangent plane intersects $G(u)$ in more than one point. I know the answer: in fact, it is on mathstackexchange somewhere and it's something like this: $v(x,y)$ being linear is also harmonic, hence so is $u-v$. If $(x_0,y_0,u(x_0,y_0))$ were to be the only intersection point, then the tangent plane is everywhere above or below the graph, then apply the maximum principle to $u-v$. However the thing I don't understand is why the tangent plane cannot be above and below the graph in spite of having one intersection point. Does this too take the maximum principle to prove? What am I missing? Many thanks in advance.
 A: We just need to fill in a little hole at the end of the logic, which is where implicitly you're referencing the intermediate value theorem.  Let's make that explicit geometrically.
Here's the entire argument:


*

*$u-v$ certainly has the one zero where it intersects the tangent plane. 

*Since $u-v$ is harmonic and attains the value $0$ at $(x_0,y_0)$, the strong maximum (and minimum) principle tells you that we can pick any neighborhood of $(x_0,y_0)$, and find points $(x_u,y_u)$ such that $(u-v)(x_u,y_u) > 0$ and $(x_b,y_b)$ such that $(u,v)(x_b,y_b) < 0$.  

*Let $\alpha(t)$ be any curve on the $(x,y)$-plane connecting $(x_b,y_b)$ to $(x_u,y_u)$ and not passing through $(x_0, y_0)$.  

*The image of $\alpha$ under $u-v$ is somewhere less than zero and somewhere greater than zero, so the intermediate value theorem says that for some $t_0$, $(u-v)(\alpha(t_0)) = 0$.

*Let $\alpha(t_0) = (x_0^*, y_0^*)$. Since $\alpha$ by construction doesn't pass through $(x_0,y_0)$, we know $(x_0^*, y_0^*) != (x_0,y_0)$. We therefore found a new zero of $u-v$, $\alpha(t_0) = (x_0^*, y_0^*)$ 


Hope this helps!
