Let \(u:\Delta\to\mathbb R\) be smooth and suppose \[ \Delta u(z)>0 \qquad(z\in\Delta). \] Show that \(u\) satisfies the maximum principle.
Solution
The strict inequality \(\Delta u>0\) rules out every interior local maximum. Consequently, on any relatively compact subdomain, the maximum of \(u\) is attained on the boundary.
1. Suppose, for contradiction, that \(u\) has a local maximum at\(z_0=(x_0,y_0)\in\Delta\).
At an interior local maximum, the restrictions \[ x\longmapsto u(x,y_0) \qquad\text{and}\qquad y\longmapsto u(x_0,y) \] each have a local maximum at \(x_0\) and \(y_0\), respectively.
2. At such a point one must have\[ u_{xx}(z_0)\le0, \qquad u_{yy}(z_0)\le0. \]
1. Smoothness of \(u\) ensures thatthese second derivatives exist.
3. Hence\[ \Delta u(z_0) =u_{xx}(z_0)+u_{yy}(z_0) \le0, \]contradicting the hypothesis \(\Delta u(z_0)>0\).
2. The result is incompatible with thestrict positivity of the Laplacian at every point of \(\Delta\).
4. Therefore \(u\) satisfies the maximum principle.
The contradiction in 3 shows that \(u\) has no interior local maximum. In particular, if \(V\Subset\Delta\) is a bounded domain on whose closure \(u\) is considered, continuity gives a maximum on \(\overline V\), and that maximum cannot occur in \(V\). Hence \[ \max_{\overline V}u =\max_{\partial V}u. \] This is the maximum principle for \(u\).