Convex functions lemma

Question

Would you help me to demonstrate the following lemma:

Consider a real-valued function which is convex on a proper open interval $(a,b)$. If $x, y, z \in (a,b)$ and $x < y < z$, then

$$ \frac{f(y)-f(x)}{y-x} \leq \frac{f(z)-f(x)}{z-x} \leq \frac{f(z)-f(y)}{z-y} $$

Thank you.

Thank you bitesizebo. I'm sorry if I've posted my question without reading the MathJax tutorial before. — Giulio, Apr 03 '18 at 09:41
I actually think the lemma is still good for not twice differentiable functions @Holo. An example could be f(x)=|x|. — Giulio, Apr 03 '18 at 14:26

score 0 · Answer 1 · answered Apr 03 '18 at 18:09

You can get an idea of the proof from a sketch. Let me prove the second equality (the first one is similar). Let $\lambda \in (0,1)$ such that $y = \lambda x + (1-\lambda)z$. By definition of convexity, $f(y) \leq \lambda f(x) + (1-\lambda)f(z)$. I will use this in the rewritten form $-\lambda f(x) \leq (1-\lambda)f(z) - f(y)$: $$\frac{f(z)-f(x)}{z-x} = \frac{\lambda(f(z)-f(x))}{\lambda(z-x)} = \frac{\lambda f(z) - \lambda f(x)}{z-(\lambda x + (1-\lambda)z)}\leq \frac{\lambda f(z)+(1-\lambda)f(z) - f(y)}{z-y}.$$

Thanks a lot @LinAlg, you've been really clear! – Giulio Apr 03 '18 at 22:09 — Giulio, Apr 03 '18 at 22:09

Convex functions lemma

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