Convexity of functions and second derivative 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?
 A: 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.
A: The following is true of a twice-differentiable real function: $f'' \geq 0$ if and only if $f$ is convex in the sense you describe.
For one direction, we can use the convexity property (the function dominates its chords) to show that $f''(x) \geq 0$ (that $f''(x) > 0$ does not obtain; any linear or affine function is convex and yet $f'' = 0$ identically).
Recall that when $f''(x)$ exists, we have
\begin{align*}
f''(x) = \lim_{h\rightarrow 0} \frac{f(x+h) + f(x-h) - 2f(x)}{h^2}
\end{align*}
Apply convexity now to $x_1 = x+h, x_2 = x-h, t = 1/2$, so that
$$
f(x+h) + f(x-h) - 2 f(x) \geq 0
$$
This inequality holds for all real $h$ and so upon dividing by $h^2$ and taking a limit, we obtain $f''(x) \geq 0$. Similarly one can show that the derivative is increasing (nondecreasing) by using the difference formula.
For the other direction, you might consider drawing a picture. You could also use the 'dominates tangents' characterization of convexity (any differentiable function $f$ is convex iff $f$ is greater than or equal to all its tangent lines at each point). Royden for instance proves all these things are equivalent in a nice "Chords Lemma" of his in Real Analysis.
