Proof involving convex function and concave function

In summary: You have just repeated the definition for a convex function, not a concave function. The ##\le## in the definition above should be ##\ge##.In summary, the set X is convex because for any ##x, y \in X## and ## t \in (0,1), f((1-t)x + ty) \le g((1-t)x + ty)##, which follows from the fact that g is concave and f is convex.
  • #1
TyroneTheDino
46
1

Homework Statement


[/B]
Let X be a vector space over ##\mathbb{R}## and ## f: X \rightarrow \mathbb{R} ## be a convex function and ##g: X \rightarrow \mathbb{R}## be a concave function. Show: The set {##x \in X: f(x) \leq g(x)##} is convex.

Homework Equations


[/B]
If f is convex ##\rightarrow## ##f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y)##
If g is concave ##\rightarrow## ##g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)##
##\forall x,y \in X ## and ## \alpha \in \mathbb{R}## where ## 0 < \alpha < 1##

The Attempt at a Solution



I just assumed that because
##f(x) \leq g(x)## ##\rightarrow## ##f(\alpha x + (1-\alpha)y) \leq g(\alpha x + (1-\alpha)y)##.
And also
##\alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y)##.

Then because we have that g is convex and f is concave from the relevant equations we have:
## f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y) \leq g(\alpha x + (1-\alpha)y)##

I know this isn't enough to show that the set is convex, but I am wondering I am on the right track of thinking and what possible next steps should be.
 
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  • #2
TyroneTheDino said:

Homework Statement


[/B]
Let X be a vector space over ##\mathbb{R}## and ## f: X \rightarrow \mathbb{R} ## be a convex function and ##g: X \rightarrow \mathbb{R}## be a concave function. Show: The set {##x \in X: f(x) \leq g(x)##} is convex.

Homework Equations


[/B]
If f is convex ##\rightarrow## ##f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y)##
If g is concave ##\rightarrow## ##g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)##
##\forall x,y \in X ## and ## \alpha \in \mathbb{R}## where ## 0 < \alpha < 1##

The Attempt at a Solution



I just assumed that because
##f(x) \leq g(x)## ##\rightarrow## ##f(\alpha x + (1-\alpha)y) \leq g(\alpha x + (1-\alpha)y)##.
And also
##\alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y)##.

Then because we have that g is convex and f is concave from the relevant equations we have:
## f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y) \leq \alpha g(x) + (1-\alpha)g(y) \leq g(\alpha x + (1-\alpha)y)##

I know this isn't enough to show that the set is convex, but I am wondering I am on the right track of thinking and what possible next steps should be.
The next step would be to write the definition of a convex set (as opposed to a convex function). Then see what you can do with your inequalities.
 
  • #3
tnich said:
The next step would be to write the definition of a convex set (as opposed to a convex function). Then see what you can do with your inequalities.

A subset E of X is called convex if, for any ##x,y \in E## and ##t \in (0,1)## then ##(1-t)x + ty \in E##.

So by the inequality I wrote since ##\alpha f(x) + (1-\alpha)f(y)## is contained in the set it is convex?
 
  • #4
TyroneTheDino said:
A subset E of X is called convex if, for any ##x,y \in E## and ##t \in (0,1)## then ##(1-t)x + ty \in E##.

So by the inequality I wrote since ##\alpha f(x) + (1-\alpha)f(y)## is contained in the set it is convex?
That is kind of the idea, but the set is two-dimensional. You just need to put together the chain of inequalities that shows that for two ordered pairs ##(x_1,y_1), (x_2,y_2)## such that ##x_1,x_2 \in X, f(x_1) \le y_1 \le g(x_1)## and ##f(x_2) \le y_2 \le g(x_2)## if follows that ##(1-t)(x_1,y_1) +t(x_2,y_2)## satisfies the same conditions for ##0\le t\le 1##.
 
  • #5
tnich said:
That is kind of the idea, but the set is two-dimensional. You just need to put together the chain of inequalities that shows that for two ordered pairs ##(x_1,y_1), (x_2,y_2)## such that ##x_1,x_2 \in X, f(x_1) \le y_1 \le g(x_1)## and ##f(x_2) \le y_2 \le g(x_2)## if follows that ##(1-t)(x_1,y_1) +t(x_2,y_2)## satisfies the same conditions for ##0\le t\le 1##.
Oops. Misread the problem. I'm wrong about the two dimensions. It does say to show that the set X is convex.
 
  • #6
tnich said:
Oops. Misread the problem. I'm wrong about the two dimensions. It does say to show that the set X is convex.
So what you need to show is that for all ##x, y \in X## and ## t \in (0,1), f((1-t)x + ty) \le g((1-t)x + ty)##.
 
  • #7
TyroneTheDino said:
If g is concave ##\rightarrow## ##g(\alpha x + (1-\alpha)y) \leq \alpha g(x) + (1-\alpha)g(y)##
##\forall x,y \in X ## and ## \alpha \in \mathbb{R}## where ## 0 < \alpha < 1##
You have just repeated the definition for a convex function, not a concave function. The ##\le## in the definition above should be ##\ge##.
 

Related to Proof involving convex function and concave function

1. What is a convex function?

A convex function is a mathematical function with a graph that curves upwards, resembling a bowl shape. This means that any two points on the graph lie above the line connecting them. The function has a positive second derivative, meaning that the rate of change of the function is increasing.

2. What is a concave function?

A concave function is a mathematical function with a graph that curves downwards, resembling a valley shape. This means that any two points on the graph lie below the line connecting them. The function has a negative second derivative, meaning that the rate of change of the function is decreasing.

3. How do you prove that a function is convex?

To prove that a function is convex, you can use the definition of convexity, which states that for any two points on the graph of the function, the value of the function evaluated at the midpoint of those two points must be less than or equal to the average of the function values at the two points. This can also be shown using the second derivative test, where a positive second derivative indicates convexity.

4. Can a function be both convex and concave?

No, a function cannot be both convex and concave. This is because the definitions of convexity and concavity are contradictory - convexity requires a positive second derivative, while concavity requires a negative second derivative.

5. What are some real-life applications of convex and concave functions?

Convex and concave functions have many real-life applications in fields such as economics, engineering, and statistics. Some examples include cost-benefit analysis, optimization problems, risk management, and utility functions. They are also commonly used in machine learning and data analysis to model and analyze data.

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