I would like to show that the following definitions of convexity for a function $f:(a,b)\to\mathbb{R}$ are equivalent:
- For all $x,y\in(a,b)$ and all $t\in(0,1)$, $f\big(tx+(1-t)y\big)\leq tf(x)+(1-t)f(y)$;
- For all $x,y,z\in(a,b)$ with $x<y<z$, $f(y)(z-x)\leq f(z)(y-x)+f(x)(z-y);$
- For all $x,y,z\in(a,b)$ with $x<y<z$: $$\frac{f(y)-f(x)}{y-x}\leq\frac{f(z)-f(y)}{z-y}$$
I have been able to show that $2\iff 3$ by multiplying by $(y-x)(z-y)$ and rearranging terms. However, I'm not sure how to show that the first definition is equivalent to the second or third. I've tried rewriting variables in terms of a parameter $t\in(0,1)$ and attempting to obtain the desired inequality via the first definition of convexity, but without any luck.
Could somebody provide a hint as to how to move from the first definition to the second or third? I'm sure there is a simple trick that I am missing but I have not been able to find it.
Let $x,y,z\in(a,b)$, and fix $t=(y-x)/(z-x)$. Consider $f\big(tz+(1-t)x\big)=f(y)$: $$f\big(tz+(1-t)x\big)\leq\frac{y-x}{z-x}f(z)+\frac{z-y}{z-x}f(x)$$ $$\implies f(y)(z-x)\leq f(z)(y-x)+f(x)(z-y)$$ This shows that $1\implies 2$. For the other direction, simply reverse the steps of the inequality.