Convex Functions with more than two terms

Question

A convex function satisfies $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$ for all $0\leq t\leq 1$. Then does this also imply that

$$f(t_1x_1+t_2x_2+\cdots+ t_kx_k)\leq t_1f(x_1)+t_2f(x_2)+\cdots +t_kf(x_k)$$ whenever $\sum_i t_i=1$ with $t_1, t_2, \ldots, t_k\geq 0$ ?

https://en.wikipedia.org/wiki/Jensen%27s_inequality#Proof_1_(finite_form) — Martin R, Jun 23 '22 at 18:59
Duplicate of https://math.stackexchange.com/q/1278761/42969 and https://math.stackexchange.com/q/1804505/42969. — Martin R, Jun 23 '22 at 19:00
Does this answer your question? Prove $f(\sum^k_{i=1} \alpha_i x_i) \leq \sum^k_{i=1} \alpha_i f(x_i) $ for a convex function f — K.defaoite, Jun 24 '22 at 19:23

score 2 · Answer 1 · answered Jun 23 '22 at 18:56

2

Yes! It does. The proof is by induction over $k\in\mathbb{N}$, the base case is just the definiton of convexity. Try writin it.
Hint: $$t_1x_1+\cdots+t_kx_k=t_1x_1+(1-t_1)[\frac{t_2}{1-t_1}x_2\cdots+\frac{t_k}{1-t_1}x_k]$$

answered Jun 23 '22 at 18:56

Guillermo García Sáez

8,159

Thanks! good idea ill give this a try – Andrew Jun 23 '22 at 18:58

Convex Functions with more than two terms

1 Answers1