Proof involving convex function and concave function

[TyroneTheDino]

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.

 

[tnich]

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.

 

[TyroneTheDino]

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?

 

[tnich]

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 $0\le t\le 1$.

 

[tnich]

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.

 

[tnich]

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 $t \in (0,1), f((1-t)x + ty) \le g((1-t)x + ty)$.

 

[Mark44]

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 $\ge$.