Concavity and convexity

Rushil

Given that $\frac {f(x_1) +\ f(x_2)}{2} \geq f( \frac{x_1 + x_2}{2} )$ , prove that

$\sum _{i=1} ^ n \frac{f(x_{i})}{n} \geq f( \frac {\sum _{i=1} ^n x_i}{n} )$

I know that this is a property of concave up functions. But can somebody prove it from the basics. I.e. Can somebody first prove the concavity/convexity and establish the result without using any theorm beyond Lagrange's Mean Value Theorm. I am preparing for IIT - JEE and this question is from there ( for special case of $n =\ 3$ ). There , one is expected to solve without using any theorm from advanced maths

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**Posted:** Thu Jul 28, 2005 7:01 am

enescu

Here's the "special case n=3":

adding up $\frac {f(x_1)+f(x_2)}{2}\ge f\left( \frac{x_1+x_2}{2}\right)$ and $\frac {f(x_3)+f(x_4)}{2}\ge f\left( \frac{x_3+x_4}{2}\right)$ yield

$\frac {f(x_1)+f(x_2)+f(x_3)+f(x_4)}{4}\ge \frac {f\left( \frac{x_1+x_2}{2}\right)+f\left( \frac{x_3+x_4}{2}\right)}{2}\ge f\l...$ .
Now, let $x_4=\frac {x_1+x_2+x_3}{3}$ . Plugging this in the previous inequality immediately gives
$\frac {f(x_1)+f(x_2)+f(x_3)}{3}\ge f\left(\frac {x_1+x_2+x_3}{3}\right)$ ,
as desired. The same trick works in the general case...Try it!

**Posted:** Thu Jul 28, 2005 12:38 pm

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Bogdan Enescu

Rafal

there is a simple inductive proof of the inequality but I'm not sure if it's so important to mention it here.

**Posted:** Thu Jul 28, 2005 1:40 pm

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out of time

Rushil

**Posted:** Fri Jul 29, 2005 2:01 am

Rafal

just look what enescu did Wink

,
1. First prove (what's easy) that for $n=2^m$ it's true.
2. Then prove that if it's true for $k+1$ then it's true for $k$ (putting $x_{k+1}=\frac{x_1+...x_k}{k}$ ).

**Posted:** Fri Jul 29, 2005 10:23 am

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out of time

Rushil

I understood the part of proving the inequality. But what about the relation of such functions to convexity????? Sad

**Posted:** Sat Jul 30, 2005 5:02 am

Arne

This is obviously the Jensen inequality for convex/concave functions...

**Posted:** Sat Jul 30, 2005 5:08 am

Rushil

exactly.

Basically I am asking for the proof of jensen's Inequality and its converse(if true)

If it's not true , I am also asking for a proof of that

**Posted:** Sat Jul 30, 2005 5:18 am

lomos_lupin

Rushil
Exactly ,what do you want? Confused

Rafal had showed a basic proof with induction.

**Posted:** Sat Aug 06, 2005 5:17 pm

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Rushil

Basically I am asking for the proof that if $f'(x) > 0$ , prove that
$\frac{ f (x_1) + f(x_2) }{2} \geq f ( \frac{x_1 + x_2}{2} )$ .

The general case can easily be obtained as you have shown above.. ,once we establish the inequality for $n = 2$ . My question is this!! To relate the inequality for $n=2$ with concavity without using graphs!! A rigorous proof plz!

**Posted:** Sat Aug 06, 2005 9:24 pm

blahblahblah

A function satisfying the above midpoint inequality is not necessarily convex.

As for a proof, one can be found here:

http://www-math.mit.edu/~rbm/18.100-F02.HMW/HMW7s.pdf

(it contains the solution to question 5.14 out of Rudin)

**Posted:** Sat Aug 06, 2005 9:32 pm

Rushil

i'm basically asking for a proof of Jensen's inequality!!!

**Posted:** Sat Aug 06, 2005 9:37 pm

blahblahblah

Well, I don't have a document with a proof of it, and I'm not going to take the time to write it down. The pdf that I gave you contains the proof that if $f$ is twice differentiable, then it is convex iff $f''(x)>0$ .

Honestly, it's not that hard (induction on the number of $\lambda_i$ 's), and in general, you should stop asking people incessantly for hints and whatnot. You write a huge number of posts with no real content, and it shouldn't really continue. Rafal gave you more than enough information to prove the result you're looking for here, and I have written the proof for

$\frac {f(x)+f(y)}{2}\geq f\left(\frac {x+y}{2}\right)\implies$

$\frac {\sum_i f(x_i)}{n}\geq f\left(\frac {\sum_i x_i}{n}\right)$

on this forum on at least three different occasions.

**Posted:** Sat Aug 06, 2005 9:44 pm

Rushil

I think I get it now. I am sorry if I posted useless posts. I wasn't clear about the definitons previously , and that is why I was confused.

The definiton of convexity IS that $\frac{ f(x_1) + f( x_2) }{2} \geq f ( \frac{x_1 + x_2}{2} )$ as aspecial case and a general form also!!!Ok now ,everything is foine and all the proofs are ok.

Obviously , the general case is easy once we establish tge above

Again, Thanks!!! Please correct me if I am wrong!!!

Soarer

**Posted:** Sat Aug 06, 2005 10:02 pm

pardesi

**Posted:** Sat Sep 29, 2007 5:50 pm

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