# Why is log M(x) convex (from Three Lines Theorem)

In the statement of the Three Lines Theorem we want to show for some function $$M:[0,1] \rightarrow \mathbb{R}$$ that $$\log M$$ is a convex function on $$[0,1]$$.

Recall:

$$f[a,b] \rightarrow \mathbb{R}$$ is convex if for all $$x,y \in [a,b]$$ and $$\lambda \in [0,1]$$ we have $$f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda)f(y)$$

The proof of the theorem proceeds to show that for any $$0 \leq s < t \leq 1$$ we have $$M(x)^{t-s} \leq M(s)^{t-x}M(t)^{x-s}.$$

How does this imply that $$\log M$$ is convex according to above definition? If I apply log on both sides I only get this $$(t-s)\log(M(x)) \leq (t-x)\log M(s) + (x-s)\log M(t).$$

• noticing $(t-x)/(t-s) + (x-s)/(t-s) = 1$ is a good start Aug 11, 2019 at 15:17

$$(t-s) f(x) \le (t-x) f(s) + (x-s) f(t)$$ for $$s < x < t$$ is an equivalent condition for convexity.
For $$x = \lambda s + (1-\lambda)t$$ you have $$t-x = \lambda (t-s) \, , \quad x-s = (1 - \lambda) (t-s) \, ,$$ so that $$(t-s)\log M(x) \leq (t-x)\log M(s) + (x-s)\log M(t)$$ becomes $$\log M(\lambda s + (1-\lambda)t) \le \lambda \log M(s) + (1-\lambda) \log M(t) \, .$$