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 April 2nd, 2018, 10:25 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 How can I show that f(a) = f(b)? Show that if f(x) is increasing and concave up on [a,b], then f((a+b)/2) < (f(a) + f(a))/2. My attempt: I would use the Mean Value Theorem. Since the function is increasing and concave up on the given interval, f((a+b)/2) < f(b). By the Mean Value Theorm, there exists a value c in the interval such that f ' (c) = (f(b) - f(a))/(b-a) > 0 since the function is increasing. That means, f (a) < f(b). So, f((a+b)/2) < f(b) = f(b)/2 + f(b)/2. I'm stuck. Since f(a) < f(b), I don't know how to complete the proof. Please help me. Thanks a lot.
April 2nd, 2018, 10:40 PM   #2
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Quote:
 Originally Posted by davedave f((a+b)/2) < (f(a) + f(a))/2.
Is that copied correctly? The rhs is f(a). Also you probably need to use the definition of concavity at some point.

 April 3rd, 2018, 04:29 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I presume you mean that f((a+b)/2)< (f(a)+ f(b))/2. You assume, without saying it explicitly, that f is differentiable and that b> a. Are those given as part of the problem?
 April 3rd, 2018, 12:48 PM #4 Global Moderator   Joined: May 2007 Posts: 6,788 Thanks: 708 The inequality comes from the definition of concave upward. You need to use a mathematical definition of this term. Thanks from topsquark
April 3rd, 2018, 06:51 PM   #5
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 Originally Posted by davedave Show that if f(x) is increasing and concave up on [a,b], then f((a+b)/2) < (f(a) + f(a))/2.
This question makes no sense. There is nothing to show. This is literally the definition of concave up (usually called convex).

https://en.wikipedia.org/wiki/Convex_function

April 5th, 2018, 10:00 AM   #6
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 Originally Posted by SDK This question makes no sense. There is nothing to show. This is literally the definition of concave up (usually called convex). https://en.wikipedia.org/wiki/Convex_function
Someone in my math club proved it. But, I don't remember how he did it.

 April 5th, 2018, 11:41 AM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Proved what? That every convex function is convex?
April 5th, 2018, 01:50 PM   #8
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 Originally Posted by SDK Proved what? That every convex function is convex?
"it" refers to the inequality as stated in the hypothesis. That is, if ..... then

prove that f((a+b)/2) < f(a)/2 + f(b)/2.

One of my club members proved this hypothesis elegantly. He won't be able to attend the club for two months because he'll be on vacation. Anyways, I'll ask him for his proof when I see him again. I was just eager to know how a proof would be done for this one. I guess I'll just wait for him to come back.

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