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First suppose $u$ is a classical harmonic function. At a local maximum of $u-v$, its Hessian matrix is negative semidefinite, so $0=\Delta u(x_0)\le \Delta v(x_0)$. At a local minimum the Hessian inequality reverses and gives $\Delta v(x_0)\le 0$. Thus both test conditions hold.

Conversely, suppose only that $u$ is continuous and satisfies the two conditions. We prove the stronger conclusion that these conditions themselves imply $u\in C^{\infty }$ and $\Delta u=0$. Use the standard Dirichlet existence theorem on a ball: every continuous function on its boundary has a harmonic extension $h\in C^{\infty }(B_R(a))\cap C(\overline{B_R(a)})$ with exactly those boundary values. This follows, for example, from the Poisson integral. Apply it to $u$ on an arbitrary relatively compact ball, obtaining a harmonic replacement $h$.

If $u-h$ is positive somewhere, choose $\epsilon >0$ small enough that $u-h-\epsilon (R^2-|x-a{|}^2)$ is still positive there. It vanishes on the boundary and therefore has a positive maximum at an interior point. The upper testhas $\Delta v=-2n\epsilon <0$, contradicting the maximum test condition. Similarly, if $u-h$ is negative somewhere, $v=h-\epsilon (R^2-|x-a{|}^2)$ yields an interior negative minimum of $u-v$ for small $\epsilon$, but $\Delta v=2n\epsilon >0$, contradicting the minimum condition. Hence $u=h$ throughout the ball.

If the quantifier requires $v\in C^2(\Omega )$ rather than a local test, multiply the displayed smooth $v$ by a smooth cutoff supported within the ball and equal to one near the touching point, extending by zero outside. The resulting global test has the same germ and Laplacian there, so the argument remains valid. Since every point lies in such a ball, $\boxed{u\text{ is a smooth harmonic function throughout }\Omega }$. This is the continuous viscosity harmonic functions are classical characterization.