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Proof involving convex function and concave function

  • #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.
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336

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
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
tnich
Homework Helper
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336
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
Homework Helper
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336
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
Homework Helper
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336
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
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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##.
 

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