Ytukyg

I’m trying to use the following definition to show the result listed below.

Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+frac{f(b)-f(a)}{b-a}(x-a)ge f(x)$$ for each $xin(a,b)$.

Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain.

Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $xin(x_1,x_2)$, $$f(x_1)+frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)ge f(x)$$

Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $xin(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$

The Mean Value Theorem guarantees the existence of $cin(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$

Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$

So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$

This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously.

Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity. For $x < y < z$ we have

$$frac{f(z) - f(y)}{z-y} > frac{f(z) - f(x)}{z-x} > frac{f(y) - f(x)}{y-x}$$

A convex function on an open interval always has one-sided derivatives, so

$$frac{f(z) - f(x)}{z-x} geqslant lim_{y to x+}frac{f(y) - f(x)}{y-x} =f'_R(x),$$

giving you half of what you need.

If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that

$$f(z) geqslant f(x) + f'(x)(z-x)$$

Fix $x_0in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $yin (a,b)$ is another point. If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider

$f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)le f'(c):$

First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities

$frac{f(d)-f(x_0)}{d-x_0}lefrac{f(c)-f(x_0)}{c-x_0}lefrac{f(c)-f(d)}{c-d}.$

Letting $dto c^-$ we see that $frac{f(d)-f(x_0)}{d-x_0}le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$.

$f'(c)$ is now a fixed number, so we can let $dto x_0^+$ in $frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)le f'(c), $ as desired.