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Let $D=[-2,2]$ and $f:D\rightarrow \mathbb{R}$ be $f(x)=4-x^2$. Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt: $f(x)=4-x^2$ is a down-facing parabola with origin at $(0,4)$. I know that. But what $D=[-2,2]$ is given for. Is it domain or a point?

Then, how do I prove that $f(x)$ is concave using the definition of a concave function? I got the inequality which should hold for $f(x)$ to be concave:

For two distinct non-negative values of $x (u$ and $v$)

$f(u)=4-u^2$ and $f(v)=4-v^2$

Condition of a concave function:

$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$

I do not know what to do next.

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$D=[-2,2]$ is a domain consisting of an interval from $-2$ through to $2$ –  Henry Apr 22 '12 at 23:20
    
Your condition for a concave (not convex) function is missing a $-$ and has a $-$ where it should have a $+$. It should be $\lambda(4-u^2)+(1-\lambda)(4-v^2)\leq 4 - [(\lambda u+(1-\lambda)v]^2$ and requires $0 \le \lambda \le 1$. –  Henry Apr 22 '12 at 23:23
    
ok I corrected the respective typos. –  Koba Apr 23 '12 at 5:05
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3 Answers

If you expand your inequality, and fiddle around you can end up with

$$ (\lambda u-\lambda v)^2\leq (\sqrt{\lambda}u-\sqrt{\lambda}v)^2. $$

Without loss of generality, you may assume that $u\geq v$. This allows you to drop the squares. Another manipulation gives you something fairly obvious. Now, work your steps backwards to give a valid proof.

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thanks but I would really appreciate if you told me what that obvious manipulation was right away because it took long time to figure it out. –  Koba Apr 23 '12 at 5:06
    
@Dostre: There are a couple of things you could simplify this to. I just moved all the $v$'s to one side and all the $u$'s to the other and factored. The assumption that $u\geq v$ gives you what you want then. –  Joe Johnson 126 Apr 23 '12 at 11:05
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up vote 2 down vote accepted

To be concave $f(x)=4-x^2$ should satisfy the condition for a concave function:

For two distinct values of x (u and v) such that $f(u)=4-u^2$ and $f(v)=4-v^2$ the following inequality should be true:

$$\lambda f(u)+(1-\lambda)f(v)\leq f(\lambda u+(1-\lambda)v), \;\text{for}\; 0<\lambda<1$$

which turns into the inequality below:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2 $$

To show that the above inequality is true, first, I expanded it and made it look like the expression Joe posted(remember that $0<\lambda<1$ by definition of a concave function). After expanding LHS I got:

$4\lambda -\lambda u^2 +4-4\lambda -v^2+\lambda v^2\leq4-[(\lambda u+(1-\lambda)v]^2$

Then, I canceled out the terms $4\lambda , 4$ and subtracted the RHS from the LHS:

$-\lambda u^2-v^2+\lambda v^2+(\lambda+v-\lambda v)^2\leq0$

After expanding the expression in parenthesis the following terms cancel out: $v^2,\lambda v^2$. Rearranging the remaining terms I got:

$\lambda^2 u^2-2\lambda^2 uv+\lambda^2 v^2\leq\lambda u^2-2\lambda uv+\lambda v^2$

Which turns into the expression Joe posted:

$(\lambda u-\lambda v)^2\leq (\sqrt{\lambda}u-\sqrt{\lambda}v)^2$

Then, I factored out $\lambda 's$ and subtracted the RHS from the LHS:

$\lambda ^2(u-v)^2-\lambda (u-v)^2\leq0$

Finally I factored out $\lambda (u-v)^2$ and got:

$\lambda (u-v)^2(1-\lambda)\leq0$

which is definitely true because by definition $0<\lambda<1$ which makes the LHS strictly negative:

$\lambda (u-v)^2(1-\lambda)<0$

I fiddled around with the terms of the inequality that should be proved for the $f(x)=4-x^2$ to be concave and turned it into the form that clearly shows that it is true for $0<\lambda<1$ which is a part of the condition for a concave function.

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If you know that for continuous functions an equivalent definition concavity is $$f\left(\frac{x+y}{2}\right) \geq \frac{f(x)+f(y)}{2}$$ and you're allowed to use this fact, the proof is easy. (This fact was mentioned in several questions at this site, perhaps more often in the dual formulation for convex functions, e.g. here.)

The following inequalities are equivalent $$\begin{align*} f\left(\frac{x+y}{2}\right) &\geq \frac{f(x)+f(y)}{2}\\ 4-\left(\frac{x+y}2\right)^2 &\ge 4-\frac{x^2+y^2}2\\ \frac{x^2+y^2}2 &\ge \left(\frac{x+y}2\right)^2\\ \frac{x^2+y^2}2 &\ge \frac{(x+y)^2}4\\ 2(x^2+y^2) &\ge (x+y)^2\\ x^2-2xy+y^2 &\ge 0\\ (x-y)^2 &\ge 0 \end{align*}$$

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I did not know about that equivalent definition. Makes the whole thing much easier. thanks. –  Koba Apr 23 '12 at 6:35
    
They are not equivalent for arbitrary functions, but if the function is continuous (which is your case), then they are. (Just wanted to remind you not to forget, that you the assumption that $f$ is continuous cannot be completely left out.) –  Martin Sleziak Apr 23 '12 at 7:47
    
Thanks very helpful –  Koba Apr 23 '12 at 7:49
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I should also have mentioned, that an example showing that the two definitions are not equivalent without any additional conditions on $f$ can be found here. –  Martin Sleziak May 1 '12 at 14:35
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