Why is the affine function $f(x) = a^T x + b$ log-concave on $\{x | a^Tx + b \gt 0\}$?

Question

Why is the affine function $f(x) = a^T x + b$ log-concave on $\{x | a^Tx + b \gt 0\}$?

I see that for it to be log-concave we must have: $$f(\theta x + (1-\theta)y) \ge f(x)^{\theta}f(y)^{1-\theta}$$ Therefore we must have $$a^T(\theta x + (1-\theta)y) + b \ge (a^Tx + b)^{\theta}(a^Ty + b)^{1-\theta}$$ Which expands to $$\theta a^T x + (1-\theta)a^Ty + b \ge (a^Tx + b)^{\theta}(a^Ty + b)^{1-\theta}$$

But I can't see how to show this inequality from here.

Anyone have any ideas?

Answer 1

The inequality you are trying to prove follows by Young’s inequality (see https://en.wikipedia.org/wiki/Young%27s_inequality_for_products), which states that if $u,v \geq 0$ and $p, q \geq 1$ with $\frac{1}{p} + \frac{1}{q} =1$, then $$ uv \leq \frac{u^p}{p} + \frac{v^q}{q}. $$ In your case, you can apply the inequality for $u = (a^Tx + b)^{\theta},$ $v = (a^Ty + b)^{1-\theta}$, $p = \frac{1}{\theta},$ and $q = \frac{1}{1 - \theta}.$

Answer 2

Proving $f(x) = a^Tx + b$ is log-concave is equivalent to proving $g(f(x)) = \log(a^Tx + b)$ is concave. Since $g(y) = \log(y)$ is concave on $y > 0$, and since composition with affine function preserves concavity, $g(f(x))$ is concave on $f(x) > 0$.