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?
Best Answer
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.