A point below the graph of a convex function lies on a line below the graph Let $\varphi:\mathbb{R}\rightarrow \mathbb{R}$ be a convex function. If $y<\varphi(x)$ why does there exist a line through $(x,y)$ which lies strictly below the graph of $\varphi$?  I ask because this is a step in the proof of Jensen's inequality for conditional expectation.
 A: One of the properties of convex functions is:

If $\varphi:(a,b) \to \mathbb{R}$ is convex, and $t_0 \in (a,b)$, there exists $\beta\in \mathbb{R}$ such that
  \begin{equation}
\varphi(x) - \varphi(t_0) \geq \beta(x-t_0)
\end{equation}
  for all $t\in(a,b)$ where $(-\infty \leq a < b \leq \infty)$.

and by rearrangement there is a line passing through $(t_0, \varphi(x))$ for all $t_0\in (a,b)$.
Let $\delta = \varphi(t_0) - y > 0$, and shift this line by $\delta$ to get the desired line.

$\varphi$ is convex if
\begin{equation}
\frac{\varphi(t)-\varphi(s)}{t-s} \leq \frac{\varphi(t')-\varphi(s')}{t'-s'}
\end{equation}
for all $s,t,s',t'\in (a,b)$ such that $s\leq s' \leq t'$ and $s< t \leq t'$.

Fix $s < t_0 < s'$. If $s' < t$, 
\begin{equation}
\frac{\varphi(s')-\varphi(s)}{s'-s} \leq \frac{\varphi(t)-\varphi(t_0)}{t-t_0}
\end{equation}
If $t<s$,
\begin{equation}
\frac{\varphi(t_0)-\varphi(t)}{t_0-t} \leq \frac{\varphi(s')-\varphi(s)}{s'-s}
\end{equation}
Choose $\beta = (\varphi(s')-\varphi(s))/(s'-s)$. As pointed out, convex functions are differentiable almost everywhere. In that case, $\beta = \varphi'(t_0)$.
A: Consider the line that passes through $(x -h, \varphi(x-h))$ and $(x +h, \varphi(x+h))$
Choose $x$ the midpoint of $x -h$ and $x+h$
Clearly the point $(x, \frac{\varphi(x+h) + \varphi(x-h)}{2})$ is on the line passing through $(x -h, \varphi(x-h))$ and $(x +h, \varphi(x+h))$.
Therefore equation of this line is:
$$y_1(x_1) = \frac{\varphi(x+h) - \varphi(x-h)}{2h}(x_1 - x) + \frac{\varphi(x+h) + \varphi(x-h)}{2}$$
By point slope from high school algebra.
Using the definition of convexity:
$$\frac{\varphi(x+h) + \varphi(x-h)}{2} > \varphi(x)$$
$$\implies y_1(x_1) > \frac{\varphi(x+h) - \varphi(x-h)}{2h}(x_1 - x) + \phi(x)$$
If we take the limit as $h\to 0$ ($h$ is arbitrary anyway), the above relation says that:
$$y_1(x_1) > \lim_{h \to 0}\frac{\varphi(x+h) - \varphi(x-h)}{2h}(x_1 - x) + \phi(x) = \varphi'(x)(x_1 -x) + \varphi(x)$$
What's remarkable about this is that the above is exactly equation of the line tangent to $\varphi$ at $x$. This means that the tangent line to $\varphi(x)$ at $x$ is below $\varphi(x^1)$ for all $x^1 \not = x$ since we can always choose $h$ such that $x + h = x^1$ and $\varphi(x^1)$ lies on a line that is above the tangent line to $\varphi(x)$ at $x^1$. 
Suppose now that we run a line through $(x, y)$ with slope $\varphi'(x)$, that line has equation:
$$y_2(x_2) = \varphi'(x)(x_2 - x) + y$$
Because $y < \varphi(x)$ and from the work we did above, 
$$\varphi(x_2) > y_1(x_2) > y_2(x_2)$$
for all $x_2$.
$$$$
Thus, we've constructed a line $y_2(x_2)$ that passes through $y$ such that $y_2(x_2) < \varphi(x_2)$ for all $x_2$.
