Is Every Integrable Function Convex and Visa-Versa?

Question

Is every integrable function also convex? Or every convex function is also integrable? Also, what is the relation between continuous and convex functions?

Answer 1

I’m going to consider the elementary case of a real single variable function $f: [a,b] \to \mathbb{R}$.

• Not every (Riemann or Lebesgue) integrable function $f$ is necessarily convex over $[a,b]$.

  For example, the function $f(x) := \begin{cases} 1 & \text{, if } x = a \\ 0 & \text{, if } a < x < b \\ -1 & \text{, if } x = b \end{cases}$ is (Riemann and Lebesgue) integrable over $[a,b]$, but it is neither convex nor concave over $[a,b]$.

• Not every convex function $f$ is continuous in $[a,b]$.

  For example, the function $f(x):= \begin{cases} 1 & \text{, if } x = a,b \\ 0 & \text{, otherwise} \end{cases}$ is convex over $[a,b]$ but it is not continuous in $[a,b]$.

• Not every continuous function $f$ in $[a,b]$ is convex over $[a,b]$.

  For example, $f(x) := (x - \frac{a+b}{2})^3$ is continuous in $[a,b]$ but it is neither convex nor concave over $[a,b]$.

But…

• Every convex function $f$ over $[a,b]$ is (Riemann and hence also Lebesgue) integrable over $[a,b]$.

  In fact, a convex function $f$ is continuous in $]a,b[$ (in general, continuity at the endpoints is missing) and bounded over $[a,b]$, therefore it is integrable by Vitali - Lebesgue Theorem.

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If $f$ is defined on a noncompact interval, say $]a,b[$, things get worse because in general you cannot recover integrability from convexity assumption.

For example, $f(x) := \frac{1}{(x-a)(b-x)}$ is convex over $]a,b[$ but it is not integrable (neither in the improper Riemann sense, nor in the Lebesgue sense) over $]a,b[$.