1

Suppose h is a continuous function on the interval $[0:1]$ such that for any $t_1, t_2 \in [0,1]$ $$ h\left(\frac{t_1 + t_2}{2}\right) \leq \frac{h(t_1) + h(t_2)}{2}$$

Show that for all natural number $n \geq 2$, and any points $t_1,t_2, \dots, t_n$ $$ h\left(\frac{t_1+t_2+\cdots + t_n}{n}\right)\leq \frac{h(t_1)+h(t_2)+\dots+h(t_n)}{n} $$

I have already attempted this problem, My approach was to use the first condition and that h is continuous to prove h is convex on the interval, and using the convexity of h to show using induction on n that the inequality holds. Now I was wondering, is there an alternative way to prove this such that I do not need to prove that h is convex?

Yadati Kiran
  • 2,320
TUFFCLUBBOY
  • 167
  • 1
    When $n=2^{m}$ we can prove this by a simple induction argument. – Kavi Rama Murthy May 03 '19 at 07:10
  • Hmm but what about for n != 2**m, seems like establishing the convexity of h seems like the only approach to this problem though... – TUFFCLUBBOY May 03 '19 at 07:17
  • 1
    Use backward induction – CY Aries May 03 '19 at 07:37
  • Oh I get it, we reasn backwards from 2**m, thanks alot. So continuity is not really needed here – TUFFCLUBBOY May 03 '19 at 07:42
  • 1
    You indeed don't need continuity for the "average-point-convexity" property you set out to prove. However, if you do not assume continuity, you cannot prove that the function is convex, unless it is also measurable - and then the proof is non-trivial. To see how full convexity is proved from your assumption, including continuity, see here:

    https://math.stackexchange.com/questions/3211912/how-do-i-prove-this-inequality-of-a-convex-function-on-an-interval

     – uniquesolution May 03 '19 at 07:52
  • @uniquesolution Please give the correct link! Your link leads me to the current question. – Ma Joad May 03 '19 at 09:15
  • https://math.stackexchange.com/questions/83383/midpoint-convex-and-continuous-implies-convex – uniquesolution May 03 '19 at 09:16

0 Answers0