---
tag: 8320TK08GR
kind: proof
name: Proof that the logarithm is strictly concave
msc: [26A51]
proves: 0ZS54B9GJF
uses: [0044CPHJBV, 07X1QA6BRW, 07ZMX4ANJS, 0BARQY52N5, 0EF2VG8XKQ]
sources:
  - url: https://proofwiki.org/wiki/Real_Function_is_Strictly_Convex_iff_Derivative_is_Strictly_Increasing
    locator: Sufficient Condition
---

## Proof

We apply the mean value theorem on two adjacent intervals.

1. Let $x \ne y$ be in $(0, \infty)$ and let $t \in (0, 1)$. If we swap $x$ with $y$ and $t$ with $1 - t$, both sides of the claimed inequality stay the same. So we may assume $x < y$.
2. Let $z = t x + (1 - t) y$. Then $z - x = (1 - t)(y - x) > 0$ and $y - z = t (y - x) > 0$. So $0 < x < z < y$.
3. The [[natural logarithm|0044CPHJBV]] is differentiable on $(0, \infty)$ with [[derivative $1/x$|07X1QA6BRW]]. So it is also [[continuous|07ZMX4ANJS]] there.
4. By the [[mean value theorem|0BARQY52N5]] on $[x, z]$, there is $c_1 \in (x, z)$ with $\ln z - \ln x = \frac{z - x}{c_1} = \frac{(1 - t)(y - x)}{c_1}$.
5. By the mean value theorem on $[z, y]$, there is $c_2 \in (z, y)$ with $\ln y - \ln z = \frac{y - z}{c_2} = \frac{t (y - x)}{c_2}$.
6. Since $0 < c_1 < z < c_2$, we have $\frac{1}{c_1} > \frac{1}{c_2}$.
7. Since $t + (1 - t) = 1$, we can write
   $$\ln z - t \ln x - (1 - t) \ln y = t (\ln z - \ln x) - (1 - t)(\ln y - \ln z).$$
8. By steps 4 and 5, the right side equals $t (1 - t)(y - x)\big(\frac{1}{c_1} - \frac{1}{c_2}\big)$.
9. Each factor is positive, by step 6 and since $0 < t < 1$ and $x < y$. So $\ln(t x + (1 - t) y) > t \ln x + (1 - t) \ln y$.
10. This holds for all $x \ne y$ in $(0, \infty)$ and all $t \in (0, 1)$. So $\ln$ is strictly [[concave|0EF2VG8XKQ]] on $(0, \infty)$.

## Idea

The point $z$ splits $[x, y]$ into two pieces. By the mean value theorem, the slope of the graph over each piece is a value of $1/x$ inside that piece. Since $1/x$ is strictly decreasing, the right slope is smaller than the left slope, so the graph bends downward and lies strictly above the chord.