$f$ convex $\iff f(y)\geq f'(x)(y-x)+f(x).$

Question

I want to prove that $f$ is convex $\iff f(y)\geq f'(x)(y-x)+f(x)$.

The implication is fine, but I have difficulties with the converse implication. I tried to prove that $f'$ is increasing, but still, it doesn't work : Let $y>x$, then $$\frac{f(y)-f(x)}{y-x}\geq f'(x),$$ but taking $x\to y$ we get $f'(y)\geq \lim_{x\to y}f'(x).$ So if $f$ is $\mathcal C^1$ we don't get better than $f'(y)\geq f'(y)$... any idea ?

Answer 1

Graphically, this means that the tangent at any point of a convex function lies below the graph.

Convexity implies $ f(y)\geq f'(x)(y-x)+f(x)$

Now, take $0 < \theta < 1$, we know that for $x,y \in \operatorname{dom}(f)$, we also have that $x + \theta(y-x) \in \operatorname{dom}(f)$. Using the definition of convexity, we can say $$f(x + \theta(y-x)) \leq (1-\theta)f(x) + \theta f(y)$$ which implies $$f(y) \geq f(x) + \frac{f(x + \theta(y-x)) -f(x)}{\theta}$$ Taking $\theta \rightarrow 0$, we get the desired result. Now, to show the other way around,

$ f(y)\geq f'(x)(y-x)+f(x)$ implies convexity

Choose any $x \neq y$, and $0 < \theta < 1$, let $z = \theta x + (1-\theta) y$. Applying the equation twice we have $$ f(x)\geq f'(z)(x-z)+f(z) \tag{1}$$ and $$ f(y)\geq f'(z)(y-z)+f(z) \tag{2}$$ Now, multiplying $(1)$ by $\theta$ and $(2)$ by $1 - \theta$, we get $\theta f(x) + (1-\theta)f(y) \geq f(z)$ which means we have convexity.

Answer 2

Choose any $x\ne y$, and $0\le\theta\le1$, and let $z=\theta x+(1-\theta)y$. Applying the inequality twice yields $$ f(x)\ge f(z)+f'(z)(x-z) \quad\text{and}\quad f(y)\ge f(z)+f'(z)(y-z). $$ Multiplying the first inequality by $\theta$, the second by $1-\theta$, and adding them yields $$ f(z)\le\theta f(x)+(1-\theta)f(y), $$ which proves that $f$ is convex.