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The two-point convexity inequality for the exponential function

Statement

For all $x, y \in R$ and $0 \leq t \leq 1$ , $\exp ((1 - t) x + t y) \leq (1 - t) \exp x + t \exp y .$ If $0 < t < 1$ , equality holds exactly when $x = y$ .

Facts & Assumptions

Given: Reals $x, y$ and $t \in [0, 1]$ .

[L1]

$\exp^{'} = \exp$ ; the chain rule differentiates compositions, and the algebra of derivatives differentiates finite linear combinations (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

The mean value theorem applies on a closed interval to a function differentiable in its interior (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L3]

The exponential is positive and strictly increasing (The exponential function is strictly increasing).

Proof

technique · direct

1.1L1L3

Put $d = y - x$ and $g (s) = 1 - s + s \exp d - \exp (s d)$ . Then $g (0) = g (1) = 0$ and $g^{''} (s) = - d^{2} \exp (s d) \leq 0$ .

2.1step 1.1L1L2

If $0 < s < 1$ , apply the mean value theorem to $g$ on $[0, s]$ and $[s, 1]$ ; since $g^{'}$ is nonincreasing by the mean value theorem applied to $g^{'}$ , their slopes give $g (s) / s \geq - g (s) / (1 - s)$ , hence $g (s) \geq 0$ .

3.1step 2.1L3algebra

At $t = 0$ , both sides of the displayed inequality equal $\exp x$ ; at $t = 1$ , both sides equal $\exp y$ . For $0 < t < 1$ , step 2.1 gives $g (t) \geq 0$ , and multiplying by $\exp x > 0$ converts this into the displayed inequality. Thus the inequality holds for every $t \in [0, 1]$ .

4.1step 1.1step 2.1L3∎

When $0 < t < 1$ and $d \neq 0$ , $g^{''} < 0$ , so the slope comparison is strict and $g (t) > 0$ ; when $d = 0$ , equality is immediate.

The two-point convexity inequality for the exponential function

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources