Given a discontinuos function, show that it is not concave

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Homework Help Overview

The problem involves a function \( f \) defined on the interval (0,1) that is discontinuous. The original poster attempts to demonstrate that such a function cannot be concave based on the definition of concavity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster discusses the need to show a specific inequality involving points of discontinuity and explores the implications of discontinuity on concavity. Other participants inquire about the context of the problem and the nature of the function being studied.

Discussion Status

The discussion is ongoing, with participants exploring definitions and implications of concavity in relation to discontinuous functions. Some guidance is offered regarding the relationship between continuity and concavity, but no consensus has been reached.

Contextual Notes

The problem is situated within a Microeconomic Theory course, and there is a mention of a textbook that covers concavity as a mathematical property, indicating a potential overlap with Real Analysis concepts.

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Homework Statement



Let f be a function from (1,0) to (1,0). Suppose that f is discontinuous. Show that f is not concave.

Homework Equations


The Attempt at a Solution

Let f:(0,1)-->(0,1). Suppose f is discontinous. Show that it is not concave.I've been working on this problem for over an hour. This is what I got so far.

What I want to show is the following:

There exists \alpha, x_{1}, x_{2} such that

\alpha f(x_{1})+(1-\alpha)f(x_{2}) \geq f(\alpha x_{1}+(1-\alpha)x_{2})

Now, let x_{1} be a point of discontinuity of f. Thus

lim_{x \rightarrow x_{1}}f(x) \neq f(x_{1})

What I'm trying to show is that we can take an epsilon-neighborhood about f(x_{1}), call it N_{\epsilon}(f(x_{1})), small enough so that for a given \alpha, such that

f(\alpha x_{1}+(1-\alpha)x_{2}) \in N_{\epsilon}(f(x_{1})), then

\alpha f(x_{1})+(1-\alpha)f(x_{2}) \geq f(\alpha x_{1}+(1-\alpha)x_{2}).

Is this correct? Can you provide any hints?

Thank you

A
 
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what kind of class is this problem from? A little context might help know where to go. I might be able to help you with showing a midpoint convex function is convex if it is continuous..
 
Hello,

It's a class in Microeconomic Theory

Thanks
 
so what you are trying to show is that if the function is not continuous, that it is convex?
 
I'm trying to show that if the function f is discontinuous, then it cannot be concave.

Here's the general definition of concavity

Let f be a function of many variables defined on the convex set S. Then f is
concave on the set S if for all x ∈ S, all x' ∈ S, and all λ ∈ (0,1) we have

f ((1−λ)x + λx') ≥ (1−λ) f (x) + λ f (x')
 
at this level of mathematics.. if you can even understand the question, you should get an "A"..

:)
 
what text are you studying from?..
 
Its called Microeconomic Theory by Mas-Collel, Winston, and Green.

I thought concavity was a general mathematical property, I learned it in Real Analysis.
 

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