Convexity on the positive reals and $𝑓 (0) \leq 0$ implies superadditivity

Let $𝑓 : [0, \infty) \to ℝ$ be a convex function and $𝑓 (0) \leq 0$ . Then #m/thm/anal

𝑓 (𝑎) + 𝑓 (𝑏) \leq 𝑓 (𝑎 + 𝑏)

and if $𝑓$ is strictly convex,

𝑓 (𝑎) + 𝑓 (𝑏) < 𝑓 (𝑎 + 𝑏)

Proof

From the definition of convexity

𝑓 (𝑡 𝑥) * \leq 𝑡 𝑓 (𝑥) + (1 - 𝑡) 𝑓 (0) \leq 𝑡 𝑓 (𝑥)

therefore

𝑓 (𝑎) + 𝑓 (𝑏) = 𝑓 ((𝑎 + 𝑏) 𝑎 𝑎 + 𝑏) + 𝑓 ((𝑎 + 𝑏) 𝑏 𝑎 + 𝑏) * \leq 𝑎 𝑎 + 𝑏 𝑓 (𝑎 + 𝑏) + 𝑏 𝑎 + 𝑏 𝑓 (𝑎 + 𝑏) = 𝑓 (𝑎 + 𝑏)

where the marked inequalities become strict for a strictly convex function.

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Convexity on the positive reals and 𝑓(0) ≤0 implies superadditivity