Which property of convex functions was used in this step?

Question

I am reading a larger proof about convex functions and came across this step, which is not at all obvious to me. Take $f$ convex and points $x_1, x_2, x_3$ such that $x_2$ is a convex combination of $x_1$ and $x_3$. Set $$\lambda=\frac{\lVert x_2-x_1\rVert}{\lVert x_3-x_1\rVert}\in(0,1),$$ then by the convexity of $f$ we have $$f(x_2)-f(x_1)\le \lambda(f(x_3)-f(x_1)) = \lVert x_2-x_1\rVert\frac{f(x_3)-f(x_1)}{\lVert x_3-x_1\rVert}.$$ This doesn't look like any transformation for convex functions I've ever seen. Can someone please tell me why the inequality is obvious?

Answer 1

It is assumed that $x_2$ lies on the line segment between $x_1$ and $x_3$. That is, there exists some $\lambda \in [0, 1]$ such that $$x_2 = \lambda x_3 + (1 - \lambda)x_1.$$ Intuitively, you can think of $\lambda$ and $1 - \lambda$ as weights for a weighted average. When $\lambda = 0$, we get back $x_1$. When $\lambda = 1$, we get back $x_3$. When $\lambda = 1/2$, we get a perfect balance between the two: the midpoint. If you want $x_2$ to be twice as far from $x_3$ as from $x_1$, then you want $\lambda = 1/3$.

So, intuitively (to me at least) it makes sense that $\lambda$ could be expressed as the proportion of the distance between $x_2$ and one of the endpoints (in this case, $x_3$), and the length of the whole interval from $x_1$ to $x_3$.

It's also super easy to prove rigorously: \begin{align*} x_2 = \lambda x_3 + (1 - \lambda)x_1 &\implies x_2 - x_1 = \lambda x_3 - \lambda x_1 \\ &\implies \|x_2 - x_1\| = \lambda\|x_3 - x_1\| \\ &\implies \lambda = \frac{\|x_2 - x_1\|}{\|x_3 - x_1\|}. \end{align*} (Note: $\lambda \ge 0$, so we don't need to put absolute values around $\lambda$.)

What they didn't make clear was the significance of $\lambda$: that it satisfies $x_2 = \lambda x_3 + (1 - \lambda)x_1$. From this, using the definition of convexity in functions, \begin{align*} f(x_2) - f(x_1) &= f(\lambda x_3 + (1 - \lambda)x_1) - f(x_1) \\ &\le \lambda f(x_3) + (1 - \lambda)f(x_1) - f(x_1) \\ &= \lambda( f(x_2) - f(x_1)) \\ &= \frac{\|x_2 - x_1\|}{\|x_3 - x_1\|}(f(x_2) - f(x_1)). \end{align*}