Prove the Riemann integrability of convex function $f:[0,1]\to \mathbb{R}$

Question

Let $f:[0,1] \to \mathbb{R}$ be convex. Recall that $f$ is convex if $$\lambda f(x) + \left(1-\lambda\right)f(y) \geq f\left(\lambda x + \left(1-\lambda\right)y\right), \forall x, y, \lambda \in [0,1]$$ Prove that $f$ is Riemann integrable on $[0,1]$.

What is the easiest way to approach this problem? We know that $f$ is continuous on $(0,1)$. Can we assume that $f$ is bounded on $[0,1]$? Then it would follow that $f$ is Riemann integrable on $(a,b)$ for all $0<a<1$ and $0<b<1$ because it is a continuous, bounded function -- and so $f$ is Riemann integrable on $[0,1]$.

If this is a valid approach, how would I prove that $f$ is bounded on $[0,1]$?

Answer 1

Let

• $L_1: \Bbb R \to \Bbb R$ be the linear function defined by $L_1(0) = f(0)$ and $L_1(1/2) = f(1/2)$,
• $L_2: \Bbb R \to \Bbb R$ be the linear function defined by $L_2(1/2) = f(1/2)$ and $L_2(1) = f(1)$,
• $L_3: \Bbb R \to \Bbb R$ be the linear function defined by $L_3(0) = f(0)$ and $L_3(1) = f(1)$.

Then

• $L_2(x) \le f(x) \le L_3(x)$ for $0 \le x \le 1/2$, and
• $L_1(x) \le f(x) \le L_3(x)$ for $1/2 \le x \le 1$,

which proves that $f$ is bounded. As you correctly said, this implies that $f$ is Riemann integrable on $[0, 1]$.

One can also show that the limits $\lim_{x \to 0^+} f(x)$ and $\lim_{x \to 1^-} f(x)$ both exist, which means that there is a continous function $F$ on $[0, 1]$ whose restriction to $(0, 1)$ coincides with $f$. This also proves that $f$ is Riemann integrable on $[0, 1]$.