Equivalent conditions for convexity without differentiability assumption

Question

Given the definition for convexity as a function $f \colon \mathbb R \to \mathbb R$ such that $$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y),$$ I am trying to show that for each fixed $b \in \mathbb R$ there exists $\beta$ such that $$ f(b) \geq f(a) + \beta(b-a)$$ for all $a \in \mathbb R$. If I knew $f$ was differentiable I would take $\beta = f'(b)$, but I do not have this additional assumption. I have noted that $$ f(b) = f(a) + f(b) - f(a) = f(a) + \frac{f(b) - f(a)}{b-a}(b-a),$$ but I do not know what to do from here. Somehow I need some kind of limiting argument, but the limit of the quotient need not exist right?

Answer 1

You are right, you have to use a limiting argument on that quotient.

Start proving that given $s < b < t$ then $$ \frac{f(b) - f(s)}{b - s} \leq \frac{f(t) - f(b)}{t - b}. $$ Take $\beta = \sup_{s < b}\frac{f(b) - f(s)}{b - s}$.

Answer 2

You will need the following lemma :

Lemma : Let $a<b<c$ be real numbers. Then $$\frac{f(b)-f(a)}{b-a}\leq \frac{f(c)-f(a)}{c-a}\leq \frac{f(c)-f(b)}{c-b}.$$

Proof : Since $a<b<c$ there must be some $0<\lambda<1$ such that $ b=\lambda a+(1-\lambda)c$. We can compute that $\lambda =\frac{c-b}{c-a}$. Then the convexity of $f$ implies that \begin{equation}f(b)\leq \lambda f(a)+(1-\lambda)f(c)= \frac{(c-b)f(a)+(b-a)f(c)}{c-a}.\end{equation} This implies that $$\frac{f(b)-f(a)}{b-a}\leq \frac{(c-b)f(a)+(b-a)f(c)-(c-a)f(a)}{(b-a)(c-a)}=\frac{f(c)-f(a)}{c-a},$$and the other inequality follows in the same way.

Now for a fixed $b$, consider the sets $$B_1=\left\{\frac{f(b)-f(a)}{b-a}\left| a<b\right.\right\}$$ $$B_2=\left\{\frac{f(c)-f(b)}{c-b}\left| b<c\right.\right\}$$ Then by the lemma, for any $\beta_1\in B_1$ and $\beta_2\in B_2$, $\beta_1\leq \beta_2$. This implies that $\sup B_1\leq \inf B_2$; now any $\beta \in [\sup B_1, \inf B_2]$ satisfies the inequality you want for all $a\in \mathbb{R}$.