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

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