In the lecture notes for a course I'm taking, the definition of a convex function is given as follows:
"a function $f$ is convex if, for any $x_1$ and $x_2$, and for any $\alpha$ $\in$ [0,1], $\alpha f(x_1) + (1-\alpha)f(x_2) \ge f(\alpha x_1 + (1-\alpha ) x_2)$"
That is, if you draw a line segment between two points on the curve for this function, the function evaluated at any $x$ between $x_1$ and $x_2$ will be lower than the line segment.
Immediately after this definition is given, there is an exercise: "show that $f(x) = -\ln(x)$ is convex." Now, I happen to know that a function is convex if its second derivative is always greater than zero, so we can easily check the second derivative here to show that $-\ln(x)$ is convex (the second derivative is $\frac{1}{x^2}$, which is always greater than zero).
However, because of the way the exercise comes immediately after the "line segment" definition of convexity, without any mention of the second derivative test, I get the impression that it is possible to prove that $-\ln(x)$ is convex without using the second derivative test. I have attacked this problem many different ways, and I haven't been able to show convexity using only the line segment definition. Is this possible?
Best Answer
It's expedient to apply the following result (see for example, Exercise $24$, Chapter $4$ of Rudin's Principles of Mathematical Analysis)
Notice that $(a, b)$ can be extended to $(0, +\infty)$.
Here, if $x, y \in (0, + \infty)$, by AM-GM inequality, $$-\ln\left(\frac{1}{2}x + \frac{1}{2}y\right) \leq -\ln(\sqrt{xy}) = \frac{1}{2}(-\ln x) + \frac{1}{2}(-\ln y).$$ Since $-\ln x$ is continuous in $(0, +\infty)$, the proof is complete.