Q:

Tangent lines for convex functions

In theorem 1 here, the author says that if $\phi$ is a convex function on $(a,b)$ then for every point $c\in (a,b)$ there exists a line $L$ that passes through $c$ such that the graph of $\phi$ lies completely above $L$. The proof is relegated to exercises. I am able to prove the inequality in the exercise that for $x<y<z$ 
$$\frac{\phi(y) - \phi(x)}{y-x} \leq \frac{\phi(z) - \phi(x)}{z-x} \leq \frac{\phi(z) - \phi(z)}{z-y} $$
I am not sure how to use this to prove the theorem. I am able to prove it in the case when $\phi$ is differentiable using Taylor's theorem, but not in general. I saw a proof here where they use monotonicity of certain sequences to show a particular limit exists, but I don't see how monotonicity alone (without boundedness) is sufficient for existence of limit.

A:

Yes, in the second link the proof is  on page $2$.  They could do away with limits, the only important thing is the following: if we have two sets of numbers 
$A$ and $B$ and $a\le b$ for every $a \in A$ and $b \in B$ then there exists $c$ so that $a \le c \le b$ for every $a\in A$ and $b \in B$.  Fix $y$. The sets are 
$$A_y = \{ \frac{f(y) - f(x)}{y-x} | x < y\}\\
 B_y = \{ \frac{f(z) - f(y)}{z-y} | z > y\}$$
So take $c_y$ in between $A_y$ and $B_y$. Then we have 
$$f(y) - f(x) \le c_y ( y - x) \textrm{ for } x < y \\
f(z) - f(y) \ge c_y ( z-y)  \textrm{ for } z > y $$ 
hence
$$f(t) \ge c_y(t-y) + f(y)$$ for all $t$. 

