A bivariate function $C(u,v)$ that maps $[0,1]^{2}$ to $[0,1]$ is a copula if it satisfies the following two conditions:
(i) Boundary conditions: \begin{align*} C(u,0) = 0\\ C(0,v) = 0\\ C(u,1) = u\\ C(1,v) = v\\ C(1,1) = 1 \end{align*}
(ii) 2-increasing property \begin{align*} C(u_{1},v_{1}) + C(u_{2},v_{2}) - C(u_{2},v_{1}) - C(u_{1},v_{2}) \geq0 \end{align*}
Based on this, how can we prove the second condition is equivalent to $$\frac{\partial^{2}C(u,v)}{\partial u\partial v}\geq 0$$
when $C(u,v)$ is twice differentiable? Please, provide a full answer if it is possible. Thanks!
Let's suppose $C(u_1,v_1) + C(u_2,v_2) - C(u_1,v_2) - C(u_2,v_1) \geq 0$, then by Mean Value Theorem $\exists\eta_1\in[0,1]:C(u_1,v_1) - C(u_2,v_1)=\frac{\partial C(\eta_1,v1)}{\partial u}$
Equally, $\exists\eta_2\in[0,1]:C(u_1,v_2) - C(u_2,v_2)=\frac{\partial C(\eta_2,v_2)}{\partial v}$
Which yields $C(u_1,v_1) + C(u_2,v_2) - C(u_1,v_2) - C(u_2,v_1) \geq 0\implies \frac{\partial C(\eta_1,v_1)}{\partial v}-\frac{\partial C(\eta_2,v_2)}{\partial v}\geq 0$
By repeating the Mean Value Theorem application as before(and using twice differentiability) $\exists\theta\in[0,1]^2:0\leq\frac{\partial C(\eta_1,v_2)}{\partial v}-\frac{\partial C(\eta_2,v_2)}{\partial v} =\frac{\partial^2 C(\theta)}{\partial u \partial v}$(Since$[0,1]^2$ is convex)
The other implication is analogous, but proceeding in reverse order