A function f is convex if it satisfies the inequality $f(tx + (1 − t)x', ty + (1 − t)y') ≤ tf(x, y) + (1 − t)f(x', y')$, for all numbers 0 ≤ t ≤ 1 and all pairs of points (x, y) and (x', y') in the domain off (which we assume to be the whole of $\Bbb{R}^2$ in the rest of this problem).
Show that $f_x(x, y_o)$ is a nondecreasing function of the variable x (for $y_o$ fixed).
Show that $f_y(x_o, y)$ is a nondecreasing function of the variable y (for $x_o$ fixed).
Just not sure how to show the above! We were given a hint for $f_x$ to show that, when $x_1<x_2<x_3$, ${f(x_2, y_o)-f(x_1, y_o)\over x_2-x_1}\le{f(x_3, y_o)-f(x_2, y_o)\over x_3-x_2}$
For $x_1<x_2<x_3$: $$f(x_2, y_o)=f(x_3\frac{x_2-x_1}{x_3-x_1}+x_1\frac{x_3-x_2}{x_3-x_1}, y_o)$$ $$\le f(x_3, y_o)\frac{x_2-x_1}{x_3-x_1} + f(x_1, y_o)\frac{x_3-x_2}{x_3-x_1}$$
Rearranging that we get: $${f(x_2, y_o)-f(x_1, y_o)\over x_2-x_1}\le{f(x_3, y_o)-f(x_2, y_o)\over x_3-x_2}$$
Thus, for any $x_1<x_2<x_3<x_4$, $${f(x_2, y_o)-f(x_1, y_o)\over x_2-x_1}\le{f(x_3, y_o)-f(x_2, y_o)\over x_3-x_2}\le {f(x_4, y_o)-f(x_3, y_o)\over x_4-x_3}$$ Let $x_2\to x_1^{+}$and $x_4\to x_3^{+}$ and we get $f_x(x_1,y_0)\le f_x(x_3,y_0)$ for any $x_1\le x_3$
Similarly show it for $f_y(x_0,y)$