Convex analysis problem

Question

I have the following problem. Let $f:[a,b]\to \mathbb{R}$ be continuously convex. I have to prove that there exists $c\in (a,b)$ such that $$\frac{f(a)-f(b)}{b-a}\in \partial f(c)$$ Firstly, I'm being doubt with $\frac{f(a)-f(b)}{b-a}$ (don't ensure this one is correct, may be it is $\frac{f(b)-f(a)}{b-a}$). Secondly, I try to prove this problem by using the following proposition $$s\in \partial f(x_0) \Leftrightarrow \forall x\in \mathbb{R}, f(x)\ge f(x_0)+s(x-x_0)$$ That means for this problem, I need to find $c\in (a,b)$ such that $$f(x)\ge f(c)+\frac{f(b)-f(a)}{b-a}(x-c)$$ or $$f(x)\ge f(c)+\frac{f(a)-f(b)}{b-a}(x-c)$$ And then, I try to apply the following inequality to this one but cannot. $$f(x)\le \frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)$$ So anybody can help me?

Answer 1

What if we mimic the standard proof for the mean value theorem?

Let $g:[a,b] \to \mathbb R$ such that \begin{equation} g(t) = f(t) - M(t - a) \end{equation} where $M$ is chosen so that $g(b) = f(a)$. Note that $g$ is convex and continuous and $g(a) = g(b)$. Hence $g$ has a minimizer $c \in (a,b)$. It follows that \begin{equation} 0 \in \partial g(c) = \partial f(c) - M \end{equation} or in other words \begin{equation} M \in \partial f(c). \end{equation}

Now $g(b) = f(a)$ gives us \begin{align} &f(b) = f(a) + M(b-a) \\ \implies &\frac{f(b) - f(a)}{b-a} = M \in \partial f(c) \end{align} which is what we wanted.