A function $f: [0,1] \rightarrow \mathbb R $ is called convex, if, for any two points $x_1$ and $x_2$ in $t\in[0,1]$,
$$f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2).$$
If $f$ is differentiable, how to prove that the above definition is equivalent to $f^\prime (x) $ being a monotonically increasing function? And hence, that $f^{\prime \prime}(x) >0$ if it exists?
Best Answer
Yes. If $f:[0,1]\to \mathbb{R}$ is convex and it has derivative $f'$ on $(0,1)$, then $f'$ is (monotonically) increasing function on $(0,1)$
Lemma If $f:[0,1]\to \mathbb{R}$ is convex and $0<x<z<y<1$, then $$\frac{f(z)-f(x)}{z-x}\le \frac{f(y)-f(x)}{y-x}\le \frac{f(y)-f(z)}{y-z}$$
Proof of lemma: Since $f$ is convex, we get $$f(z)=f\left( \frac{y-z}{y-x} x + \frac{z-x}{y-x}y\right)\le \frac{y-z}{y-x}f(x)+\frac{z-x}{y-x}f(y).$$ Multiply $y-x$ both sides and transpose it properly, we get
$$(y-x)(f(z)-f(x))\le (z-x)(f(y)-f(x))$$
and divide $(y-x)(z-x)$ both sides, we get $$\frac{f(z)-f(x)}{z-x}\le \frac{f(y)-f(x)}{y-x}.$$
Likewise, we can get $$\frac{f(y)-f(x)}{y-x}\le \frac{f(y)-f(z)}{y-z}.$$
Proof of theorem: Let $0<x<y<1$, $0<x<s<z<t<y<1$. By lemma, we get $$\frac{f(s)-f(x)}{s-x}\le \frac{f(z)-f(x)}{z-x}\le \frac{f(z)-f(s)}{z-s}\quad \cdots (1)$$ $$\frac{f(z)-f(s)}{z-s}\le \frac{f(t)-f(s)}{t-s}\le \frac{f(t)-f(z)}{t-z}\quad \cdots (2)$$ $$\frac{f(t)-f(z)}{t-z}\le \frac{f(y)-f(z)}{y-z}\le \frac{f(y)-f(t)}{y-t}\quad \cdots (3)$$
Combine these inequalities, we get $$\frac{f(s)-f(x)}{s-x}\le \frac{f(z)-f(x)}{z-x} \le \frac{f(y)-f(z)}{y-z}\le \frac{f(y)-f(t)}{y-t}$$
Take $s\to x$, $t\to y$, we get $$f'(x)\le \frac{f(z)-f(x)}{z-x} \le \frac{f(y)-f(z)}{y-z} \le f'(y)$$ so $f'$ is increasing function.