From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,y\in dom f$,
$$f(y)\ge f(x)+\nabla f(x)^T(y-x).$$
I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:
They start out by picking $\theta\in[0,1]$, and $x,y\in dom f$. If $f$ is a single-variable function, the assumed equality tells us that $\theta f(x)\ge \theta f(z)+\theta f'(z)(x-z)$ and $(1-\theta)f(y)\ge (1-\theta) f(z)+(1-\theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $\theta f(x)+(1-\theta) f(y)\ge f(z)$. So it must be the case that $\theta f'(z)(x-z)+(1-\theta)f'(z)(y-z)\ge 0$. I'm not able to show this.
Any help is greatly appreciated. Thanks in advance.
Best Answer
The estimates are applied to $x, y$, and $z= \theta x + (1-\theta) y$. Adding the inequalities gives $$ \theta f(x) + (1-\theta) f(y)\ge f(z) + \bigl(\underbrace{\theta (x-z) + (1-\theta) (y-z)}_{=0}\bigr) f'(z) = f(z) \, . $$