Given a discontinuos function, show that it is not concave

In summary: Is this what you are trying to say?In summary, the author is trying to show that a discontinuous function cannot be concave. However, this is only true if the function is defined on the convex set.
  • #1
michonamona
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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 [itex]\alpha, x_{1}, x_{2}[/itex] such that

[itex]\alpha f(x_{1})+(1-\alpha)f(x_{2}) \geq f(\alpha x_{1}+(1-\alpha)x_{2}) [/itex]

Now, let [itex]x_{1}[/itex] be a point of discontinuity of f. Thus

[itex]lim_{x \rightarrow x_{1}}f(x) \neq f(x_{1}) [/itex]

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

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

[itex]\alpha f(x_{1})+(1-\alpha)f(x_{2}) \geq f(\alpha x_{1}+(1-\alpha)x_{2}) [/itex].

Is this correct? Can you provide any hints?

Thank you

A
 
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  • #2
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..
 
  • #3
Hello,

It's a class in Microeconomic Theory

Thanks
 
  • #4
so what you are trying to show is that if the function is not continuous, that it is convex?
 
  • #5
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')
 
  • #6
at this level of mathematics.. if you can even understand the question, you should get an "A"..

:)
 
  • #7
what text are you studying from?..
 
  • #8
Its called Microeconomic Theory by Mas-Collel, Winston, and Green.

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

FAQ: Given a discontinuos function, show that it is not concave

What does it mean for a function to be discontinuous?

A function is discontinuous if there is a break or gap in its graph. This means that the function is not defined or does not have a value at certain points on its domain.

How do you determine if a function is concave?

A function is concave if its graph curves downward, meaning that any line segment connecting two points on the graph will always lie below the graph itself. This can be determined by looking at the second derivative of the function, which should be negative for all points on the function's domain.

Can a discontinuous function be concave?

No, a discontinuous function cannot be concave. This is because for a function to be concave, it must have a continuous second derivative. Since a discontinuous function has breaks or gaps in its domain, its second derivative is not defined at those points and therefore cannot be continuous.

How do you show that a function is not concave?

To show that a function is not concave, you can either graph the function and see if it curves downward, or you can take the second derivative and check if it is negative for all points on the function's domain. If the second derivative is not negative at any point, then the function is not concave.

Can a function be concave on one part of its domain and not on another?

Yes, it is possible for a function to be concave on one part of its domain and not on another. This can happen if the function is discontinuous at certain points, as the concavity of a function is determined by its second derivative, which may change at discontinuities. Therefore, the function can be concave on one part of its domain and not on another due to these breaks or gaps in the function's graph.

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