# Integral of an increasing function is convex?

Let $f$ be a real valued differentiable function defined for all $x \geq a$. Consider a function F defined by $F(x) = \int_a^x f(t) dt$. If f is increasing on any interval, then on that interval F is convex.

I am not sure I intuitively understand this. What is the function is increasing at an increasing rate?

In this figure, for example, isn’t the set defined by the integral not convex? Since any line joining the two points of the set are lying outside the said set? Can someone explain what I’m missing?

#### Solutions Collecting From Web of "Integral of an increasing function is convex?"

Let $a < x_1 < x_2$, we have
$$F(x_2) – F(\frac{x_1 + x_2}{2}) = \int_{\frac{x_1+x_2}{2}}^{x_2}f(x) dx$$
and
$$F(\frac{x_1 + x_2}{2}) – F(x_1) = \int_{x_1}^{\frac{x_1 + x_2}{2}}f(x) dx$$

Since $f(x)$ is increasing, we have
$$\int_{\frac{x_1+x_2}{2}}^{x_2}f(x) dx \geq \int_{x_1}^{\frac{x_1 + x_2}{2}}f(x) dx$$
thus
$$F(x_2) – F(\frac{x_1 + x_2}{2}) \geq F(\frac{x_1 + x_2}{2}) – F(x_1)$$
implying
$$\frac{F(x_1) + F(x_2)}{2} \geq F(\frac{x_1 + x_2}{2})$$

Moreover, $F(x)$ is continuous due to the continuity of $f(x)$. Thus $F(x)$ is convex.

Hint:

A twice differentiable function is convex if its second derivative is positive.

By the Fundamental Theorem of Calculus $F'(x)=f(x)$. Since $F’$ is increasing, $F$ is convex.

Certainly the yellow set is not convex. But that has nothing to do with the convexity of the function $F$. In terms of graphs, a function $h\colon I\to \mathbb{R}$ where $I$ is an interval is convex if the set (sometimes called epigrap)
$$\{(x,y)\in\mathbb{R}^2:x\in I,\ h(x)\le y\}$$
is convex.