6
$\begingroup$

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?enter image description here

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?

$\endgroup$
3
  • 1
    $\begingroup$ The set in your figure is not the set above the graph of $F(x)$, which isn't reprensented in the figure. $\endgroup$
    – Bernard
    Commented Jun 9, 2015 at 12:17
  • $\begingroup$ Isn't the yellow shaded region what F(x) should be according to the question? $\endgroup$
    – dexter
    Commented Jun 9, 2015 at 12:18
  • 2
    $\begingroup$ 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. $\endgroup$
    – Bernard
    Commented Jun 9, 2015 at 12:36

3 Answers 3

15
$\begingroup$

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.

$\endgroup$
4
  • 3
    $\begingroup$ Needs a minor fix: midpoint convexity does not imply convexity, but midpoint convexity plus continuity does. $\endgroup$ Commented Jun 9, 2015 at 12:26
  • $\begingroup$ @JackD'Aurizio $f(x)$ is differentiable in $(a, \infty)$, so it should be continuous. $\endgroup$
    – PSPACEhard
    Commented Jun 9, 2015 at 12:31
  • $\begingroup$ (+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. $\endgroup$ Commented Jun 9, 2015 at 12:35
  • 1
    $\begingroup$ "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. $\endgroup$ Commented Mar 20, 2017 at 12:59
7
$\begingroup$

Hint:

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

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – dexter
    Commented Jun 9, 2015 at 12:22
  • $\begingroup$ @dexter The second derivative of $F$ is not the same as the second derivative of $f$. $\endgroup$
    – 5xum
    Commented Jun 9, 2015 at 12:25
  • $\begingroup$ OHHHHH. I just got it haha, thanks! $\endgroup$
    – dexter
    Commented Jun 9, 2015 at 13:08
7
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .