A C^2 function is subharmonic exactly when its Laplacian is nonnegative

Assume first that $u$ is subharmonic. Fix $a\in \mathrm{\Omega }$ and choose $r>0$ with $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For $0<\rho <r$, Taylor's formula in the direction $e^{it}$ gives [L1, given, algebra] $u(a+\rho e^{it})=u(a)+\rho \mathrm{\nabla }u(a)\text{\hspace{0.17em}⁣}\cdot \text{\hspace{0.17em}⁣}{e}^{it}+\frac{{\rho }^2}{2}{e}^{it}\text{\hspace{0.17em}⁣}{}^T(D^{2}u(a))e^{it}+o({\rho }^2)\mathrm{.}$ Averaging over $t$ kills the linear term and averages the quadratic term to $\frac{{\rho }^2}{4}\mathrm{\Delta }u(a)$, so the submean inequality yields $0\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+\rho e^{it})dt-u(a)=\frac{{\rho }^2}{4}\mathrm{\Delta }u(a)+o({\rho }^2)\mathrm{.}$ Dividing by ${\rho }^2$ and letting $\rho \downarrow 0$ gives $\mathrm{\Delta }u(a)\ge 0$.

Assume now that $\mathrm{\Delta }u\ge 0$ on $\mathrm{\Omega }$. Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$, and let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, and satisfy $h\ge u$ on $\mathrm{\partial }D(a,r)$. Put $w=u-h$. Then $w$ is continuous on $\overline{D(a,r)}$, belongs to $C^2(D(a,r))$, satisfies $\mathrm{\Delta }w=\mathrm{\Delta }u\ge 0$ on $D(a,r)$, and has $w\le 0$ on $\mathrm{\partial }D(a,r)$. If some point $p\in D(a,r)$ had $w(p)>0$, choose $0<\epsilon <w(p)\mathrm{/}{r}^2$ and define $w_{\epsilon }(z):=w(z)+\epsilon \mathrm{\mid }z-a{\mathrm{\mid }}^2$. On the boundary one has $w_{\epsilon }\le \epsilon r^2<w(p)\le w_{\epsilon }(p)$, so $w_{\epsilon }$ attains its maximum at an interior point $q\in D(a,r)$. The second-derivative test there gives $\mathrm{\Delta }{w}_{\epsilon }(q)\le 0$, but $\mathrm{\Delta }{w}_{\epsilon }=\mathrm{\Delta }w+4\epsilon \ge 4\epsilon >0,$ a contradiction. Therefore $w\le 0$ on $D(a,r)$, so $u\le h$ on the whole disc. Since the disc and the harmonic boundary majorant $h$ were arbitrary, [L1] shows that $u$ is subharmonic on $\mathrm{\Omega }$.