Equivalent definitions of convexity

Question

We defined the convexity of an $f: \mathbb{R} \to \mathbb{R}$ function like this:

$f$ is convex on the interval $I$ if $\forall a,b\in I$, $a<b$, $\forall x \in I$ $$f(x)\leqslant f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$$

And then we had the following theorem:

The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $\forall a \in I$, $m_a(x)=\frac{f(x)-f(a)}{x-a}$ is monotone increasing on $I\setminus\{a\}$

I tried to prove (i)$\implies$(ii) like this: Let $a,x,y \in I$. If $a<x<y$, then $$f(x) \leqslant f(a) + \frac{f(y)-f(a)}{y-a}(x-a)$$ $$\frac{f(x)-f(a)}{x-a} \leqslant \frac{f(y)-f(a)}{y-a}$$ $$m_a(x) \leqslant m_a(y)$$ Which is fine, but when $x<a<y$, we have that $x-a \leqslant 0$, so $$\frac{f(x)-f(a)}{x-a} \geqslant \frac{f(y)-f(a)}{y-a}$$ $$m_a(x) \geqslant m_a(y)$$ And it's not good. Did I miss something or is the theorem false as it is stated?

Answer 1

If $x<a<y$, one should use $$ f(a) \leqslant f(x) + \frac{f(y)-f(x)}{y-x}(a-x)\tag{*} $$ instead of $$ f(x) \leqslant f(a) + \frac{f(y)-f(a)}{y-a}(x-a), $$ which holds for $a<x<y$. Then one gets $$ m_a(x)\leq m_x(y). $$ One can also show that $$ m_x(y) \leq m_a(y) $$ which is equivalent to $(*)$.

Answer 2

You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $\frac{d^2}{(dx)^2}(x)\geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.

The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.