# Given a discontinuos function, show that it is not concave

## Homework Statement

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

## 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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## Answers and Replies

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.