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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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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.

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)