Convex function

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

Let 𝑓 :[0,∞) →ℝ be a convex function and 𝑓(0) ≤0. Then anal

𝑓(𝑎)+𝑓(𝑏)≤𝑓(𝑎+𝑏)

and if 𝑓 is strictly convex,

𝑓(𝑎)+𝑓(𝑏)<𝑓(𝑎+𝑏)

Proof

From the definition of convexity

𝑓(𝑡𝑥)∗≤𝑡𝑓(𝑥)+(1−𝑡)𝑓(0)≤𝑡𝑓(𝑥)

therefore

𝑓(𝑎)+𝑓(𝑏)=𝑓((𝑎+𝑏)𝑎𝑎+𝑏)+𝑓((𝑎+𝑏)𝑏𝑎+𝑏)∗≤𝑎𝑎+𝑏𝑓(𝑎+𝑏)+𝑏𝑎+𝑏𝑓(𝑎+𝑏)=𝑓(𝑎+𝑏)

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

---

develop | en | SemBr