📂Functions

Various Properties of Convex Functions

Theorem¹

1. All convex functions are continuous.

2. If $f$ is an increasing convex function, and $g$ is a convex function, then $f \circ g$ is also a convex function.

3. If $f$ is convex in $(a, b)$, and if $a \lt s \lt t \lt u \lt b$, then $$\dfrac{f(t) - f(s)}{t-s} \le \dfrac{ f(u) - f(s) }{ u - s } \le \dfrac{ f(u) - f(t) }{ u - t }$$

4. If $f$ is a continuous function defined in $(a, b)$ that satisfies the following, then it is a convex function. $$f \left( \dfrac{ x + y }{ 2 } \right) \le \dfrac{ f(x) + f(y) }{ 2 },\quad x,y \in (a,b)$$

5. Let $f=F^{\prime}$ be an increasing function. Then $F$ is convex.

Proof

5².

Let’s say $a\le x \le y \le b$.

$$\begin{align*} F(y) - F(\frac{ x + y }{ 2 }) &= \int_{\frac{ x + y }{ 2 }}^{y} f(t) dt \\ F(\frac{ x + y }{ 2 }) - F(x) &= \int_{x}^{\frac{ x + y }{ 2 }} f(t) dt \end{align*}$$

Since $f$ is an increasing function,

$$\int_{x}^{\frac{ x + y }{ 2 }} f(t) dt \le \int_{\frac{ x + y }{ 2 }}^{y} f(t) dt \ge$$

$$\implies F(\frac{ x + y }{ 2 }) - F(x) \le F(y) - F(\frac{ x + y }{ 2 })$$

$$2 \implies F(\frac{ x + y }{ 2 }) \le F(y) + F(x)$$

$$\implies F(\frac{ x + y }{ 2 }) \le \dfrac{ F(y) + F(x) }{ 2 }$$

According to 4., $F$ is convex.

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