Using the definition of a concave function prove that $f(x)=4-x^2$ is concave (do not use derivative).

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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:

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

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

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

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.