# 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?

• The set in your figure is not the set above the graph of $F(x)$, which isn't reprensented in the figure. – Bernard Jun 9 '15 at 12:17
• Isn't the yellow shaded region what F(x) should be according to the question? – dexter Jun 9 '15 at 12:18
• The area of the shaded region is the value of $F(x)$, but it corresponds to a point on the graph of $F(x)$. So you can't see on your figure whether $F$ is convex or not. – Bernard Jun 9 '15 at 12:36

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.

• Needs a minor fix: midpoint convexity does not imply convexity, but midpoint convexity plus continuity does. – Jack D'Aurizio Jun 9 '15 at 12:26
• @JackD'Aurizio $f(x)$ is differentiable in $(a, \infty)$, so it should be continuous. – PSPACEhard Jun 9 '15 at 12:31
• (+1) I totally agree that $F$ is continuous, my point was that convexity follows from midpoint-convexity (you proved that) and continuity. It should be mentioned. I just see the edit: nothing to say now. – Jack D'Aurizio Jun 9 '15 at 12:35
• "due to the continuity of $f(x)$": the continuity of $f$ is not needed. Because $f$ is increasing it is locally bounded and thus $F$ is locally Lipschitz. – Jonas Meyer Mar 20 '17 at 12:59

Hint:

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

• I don't think I get it. All we're given in the statement is that f is increasing in the interval, i.e the first derivative is positive. Does anything in the question also imply anything about the second derivative? – dexter Jun 9 '15 at 12:22
• @dexter The second derivative of $F$ is not the same as the second derivative of $f$. – 5xum Jun 9 '15 at 12:25
• OHHHHH. I just got it haha, thanks! – dexter Jun 9 '15 at 13:08

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.