Integral of an increasing function is convex?

Solution 1:

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.

Solution 2:

Hint:

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

Solution 3:

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.