I wrote this up because I was challenged to do it by Stephen Boyd during Convex Optimization. His point was that this is a really key fact in high dimensional probability but somehow there isn’t a simpler proof of this.

A function $f$ is log-concave if $f(\lambda x + (1-\lambda)y)\le f(x)^{\lambda} f(y)^{1-\lambda}$ for all $x,y$ and $\lambda\in [0,1]$.

We would like to show:

Problem. If $f(x,y)$ is continuous, measurable and log-concave (where $f:\mathbb R^m \times \mathbb R^k \to \mathbb R$, then $g(x) = \int f(x,y)dy$ is log-concave.

Proof. We will only do this for 1-dimensional $y$ (i.e. $k=1$), since otherwise we can just repeatedly integrate along single dimensions.

We first note some invariance when multiplying by exponentials of affine functions. Note that for $c\in\mathbb R, v\in \mathbb R^m$:

(1) $f(x,y)$ is log-concave iff $f(x,y)e^{c+\langle x,v\rangle}$ is log-concave

(2) $\int f(x,y) dy$ is log-concave iff $\int f(x,y)e^{c+\langle x,v\rangle} dy = e^{c+\langle x,v\rangle}\cdot \int f(x,y)dy$ is log-concave.

Fix $x_0, x_1\in \mathbb R^m, \lambda\in [0,1]$. We must show $$\int f(\lambda x_0+(1-\lambda) x_1, y) dy \le \left( \int f(x_0, y) dy \right)^\lambda \left( \int f(x_1, y) dy \right)^{1-\lambda} $$

so it is clear that the only parts of $f$ that matter are the three “slices” $$f_\lambda:=f(\lambda x_0 + (1-\lambda) x_1, \bullet),\quad f_0:= f(x_0, \bullet), \quad f_1:=f(x_1, \bullet).$$ These three slices also satisfy the inequality $$f_\lambda(\lambda y_0 + (1-\lambda )y_1) \ge f_0(y_0)^\lambda f_1(y_1)^{1-\lambda}$$ by considering log-concavity between the points $(x_0,y_0)$ and $(x_1,y_1)$.

By varying $c$ and $v$, we can force $\sup f_0 = \sup f_1$. Now we will show the stronger inequality: $$\int f_\lambda(y) dy \ge \lambda \int f_0(y) dy + (1-\lambda) \int f_1(y) dy \qquad (\star)$$ from which the original inequality is obtained via weighted AM-GM $$\lambda \int f_0(y) dy + (1-\lambda) \int f_1(y) dy \ge \left( \int f_0(y) dy \right)^\lambda \left(\int f_1(y) dy\right)^{1-\lambda}$$

$(\star)$ actually follows from the following lemma:

Lemma. Let $\mu\{\bullet \}$ denote the Lebesgue measure (total length) of a set. Then, $$\mu\{f_\lambda > c\} \ge \lambda \cdot \mu\{f_0 > c\} + (1-\lambda) \cdot \mu\{f_1 > c\}\qquad (\diamond)$$

Proof. $\{f_1(y) > c \}$, $\{f_0(y) > c\}$ and $\{ f_\lambda(y) > c\}$ are actually parallel sections of the set $\{f(x,y) > c\}$, which is a convex open set. See picture:

The last fact we need is that $$\int_0^\infty \mu \{f > c\} dc = \int f(y) dy$$

Since we assumed $\sup f_0 = \sup f_1$, both sets in the RHS of $(\diamond)$ are both empty or both non-empty. So by integrating $c$ from 0 to $\infty$ in $(\diamond)$, we obtain exactly $(\star)$.