2
$\begingroup$

Let

f : (a,b) → R is a midpoint-convex function (I didn't say continuity).

Here I'd like to verify following inequality ""directly"".

f( (x1+x2+x3)/3 ) ≤ (f(x1)+f(x2)+f(x3))/3

..

I can easily demonstrate for n=2^k

BUT not case for n=2^k, HOW can I demonstrate this inequallity?

Thanks for your consideration.

$\endgroup$
5
$\begingroup$

Assume you can do it for $n=2^k$, specifically $n=4$, then $f(\frac{(x_1+x_2+x_3+(x_1+x_2+x_3)/3)}{4})\leq \frac{f(x_1)+f(x_2)+f(x_3)+f((x_1+x_2+x_3)/3)}{4}$
rearrange to get n=3 case. Same approach can be generalized to any non $2^k$ number.

$\endgroup$
  • 1
    $\begingroup$ Thanks a lot. This is exactly what I want. (Sorry for being late check your answer.) $\endgroup$ – user143993 Sep 26 '14 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.