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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 $$(t-s)\log(M(x)) \leq (t-x)\log M(s) + (x-s)\log M(t).$$

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

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$$ (t-s) f(x) \le (t-x) f(s) + (x-s) f(t) $$ 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 $$ \log M(\lambda s + (1-\lambda)t) \le \lambda \log M(s) + (1-\lambda) \log M(t) \, . $$

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  • $\begingroup$ Thanks, I tried it myself but did something wrong when I wrote it out. Now it's clear. $\endgroup$
    – EinStone
    Aug 11, 2019 at 15:23

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