A point below the graph of a convex function lies on a line below the graph

Question

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.

Answer 1

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.

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$\varphi$ 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, $\beta = \varphi'(t_0)$.

Answer 2

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 $x_2$.

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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 $x_2$.