Show convex function is increasing in both variables of difference quotient using alternative definition of convexity

Question

Let $\phi$ be a function that satisfies $$\frac{\phi (t) - \phi (s)}{t - s} \leq \frac{\phi (u) - \phi (t)}{u - t}$$

where $s < t < u$.

Is it possible to directly use this definition of convexity to prove that $\phi$'s difference quotients are increasing in each variable, i.e., $$\frac{\phi (u) - \phi (s)}{u - s} \leq \frac{\phi (u) - \phi (t)}{u - t}$$ and $$\frac{\phi (t) - \phi (s)}{t - s} \leq \frac{\phi (u) - \phi (s)}{u - s}$$

where again $s < t < u$.

Background: $\phi$ is convex and using the usual definition of convexity the proof is fairly direct. So I'm curious if there's a direct proof from this definition of convexity.

Answer 1

All three relations are equivalent, and equivalent to $$ \begin{vmatrix} 1 & 1 & 1 \\ s & t & u \\ \phi(s) & \phi(t) & \phi(u) \end{vmatrix} \ge 0 $$ for $s < t < u$. In fact this determinant is equal to each of $$\begin{vmatrix} 1 & 0 & 0 \\ * & t-s & u-t \\ * & \phi(t) - \phi(s) & \phi(u)-\phi(t) \end{vmatrix} = ( \phi(u)-\phi(t)) (t-s) - (\phi(t) - \phi(s))(u-t) $$ $$ \begin{vmatrix} 0 & 0 & 1 \\ s -u & t - u & * \\ \phi(s) - \phi(u) & \phi(t) - \phi(u) & * \end{vmatrix} = (\phi(u) - \phi(t))(u-s) - (\phi(u) - \phi(s))(u-t) $$ $$ \begin{vmatrix} 1 & 0 & 0 \\ * & t-s & u-s \\ * & \phi(t)-\phi(s) & \phi(u)-\phi(s) \end{vmatrix} = (\phi(u)-\phi(s))(t-s)) - (\phi(t)-\phi(s))(u-s) $$ as can be seen by elementary column operations on the determinant.

Answer 2

Let $S$ be the point $(s, \phi (s))$, and define $T$ and $U$ respectively. The condition given is that the gradient of $ST$ is less than that of $TU$, where the $x$-coordinate of $T$ is between the other two. Draw the triangle $STU$, and interpret all difference quotients as gradients. The result should now be clear.