Distribution of the midrange in the general case Given an iid sample  $X_1, \cdots, X_n$ drawn from a  sufficiently nice (finite expectation, maybe L2 integrable, etc.) distribution, I want I want to know the population CDF of the mid-range or center point
$$ C(x) = \mathbb P \left[\frac {\max X_i+\min X_i}2 \leq 2x\right] = \mathbb P(\max X_i+\min X_i \leq 2x)$$
or at least some interesting facts and inequalities about it. This isn't a homework problem so nailing an answer  for an exact question isn't important. Maybe this is easy with extra assumptions? 
I will obviously report the results of my continued research here for everyone to share.
These are my goals. Let me share my best work so far.

The density $C'(x)$ has to be the convolution of the densities for $\min X_i, \max X_i$ whose CDF is respectively
$$ m(x) = [\mathbb P(X<x)]^n $$
and for symmetric reasons 
$$ M(x) = [1-\mathbb P(X<x)]^n $$
What I have in my Batman utility belt is the Laplace transform. (Edit: as per comments below, I passed over the fact that $\min X_i, \max X_i$ are not independent r.v.s so the convolution method doesn't apply. Please refer to the nice accepted answer) It's nice because $\mathcal L(M'(x)) = s\mathcal L(M)(s)$ so
$$ m'*M' = \mathcal L^{-1} \left\{[s\mathcal L(m)(s)] [s\mathcal L(M)(s)]\right\} = \mathcal L^{-1} \{ s^2 \left([\mathcal Lm](s) \cdot [\mathcal LM](s)\right) $$
and um, frantically going over the list of common facts about the Laplace transform,  we see that
$$ \mathcal L^{-1} \{s^2 [\mathcal Lf(s)] - f'(0) \} = f''(x)$$
which would be awesome if I was trying to inverse Laplace $s^2(\mathcal Lm + \mathcal LM) = s^2 \mathcal (L+M)$ rather than $s^2[\mathcal L m]\cdot [\mathcal LM]$. Which is close enough to be frustrating. 
I'm at a loss by now. Am I even going in the right direction? 
 A: As currently stated, the equation at the top of the question appears to have an extra (or missing) $2$ somewhere, so I will tackle the probability
$$
P\left(\min_i X_i + \max_i X_i \leq a\right)
$$
for an arbitrary $a \in \mathbb{R}$ that you can scale as needed for your context.

Proposition 1.
  Let $X_1, \ldots, X_n$ be an i.i.d. sample from an absolutely continuous distribution with probability density $f$ and cumulative distribution function $F$.
  Let $X_{(1)} = \min_i X_i$ and $X_{(n)} = \max_i X_i$.
  Then the joint density of $(X_{(1)}, X_{(n)})$, call it $g$, is given by
  $$
g(x, y)
= n (n - 1) \left(F(y) - F(x)\right)^{n - 2} f(x) f(y)
$$
  for all $(x, y) \in \mathbb{R}^2$ with $x \leq y$ ($g(x, y) = 0$ otherwise).

Proof.
We first compute
$$
P(X_{(1)} \geq x, X_{(n)} \leq y)
$$
for all $x, y \in \mathbb{R}$. This probability is clearly zero if $x > y$, so assume $x \leq y$.
Notice that
$$
\{X_{(1)} \geq x\}
= \{X_1 \geq x, \ldots, X_n \geq x\}
$$
and
$$
\{X_{(n)} \leq y\}
= \{X_1 \leq y, \ldots, X_n \leq y\},
$$
whence
$$
\begin{aligned}
P(X_{(1)} \geq x, X_{(n)} \leq y)
&= P(x \leq X_1 \leq y, \ldots, x \leq X_n \leq y) \\
&= P(x \leq X_1 \leq y) \cdots P(x \leq X_n \leq y) \\
&= \left(F(y) - F(x)\right)^n.
\end{aligned}
$$
Here we used the independence and identically-distributed-ness of $X_1, \ldots, X_n$ and the fact that their distributions are absolutely continuous.
To summarize,
$$
P(X_{(1)} \geq x, X_{(n)} \leq y)
= \begin{cases}
\left(F(y) - F(x)\right)^n & \text{if $x \leq y$} \\
0 & \text{otherwise.}
\end{cases}
$$
The joint density of $(X_{(1)}, X_{(n)})$ is now obtained by differentiating:
$$
g(x, y)
= -\frac{\partial^2}{\partial x \partial y} P(X_{(1)} \geq x, X_{(n)} \leq y)
$$
for a.e. $(x, y) \in \mathbb{R}^2$.
When $x > y$, both sides are zero, and when $x \leq y$, then a.e. we have
$$
\begin{aligned}
g(x, y)
&= -\frac{\partial^2}{\partial x \partial y} P(X_{(1)} \geq x, X_{(n)} \leq y) \\
&= -\frac{\partial^2}{\partial x \partial y} \left(F(y) - F(x)\right)^n \\
&= n (n - 1) \left(F(y) - F(x)\right)^{n - 2} f(x) f(y),
\end{aligned}
$$
as claimed.

Proposition 2.
  With the same assumptions and notation as in Proposition 1, we have
  $$
P(X_{(1)} + X_{(n)} \leq a)
= n (n - 1) \int_{-\infty}^{a / 2} f(x) \left(\int_x^{a - x} \left(F(y) - F(x)\right)^{n - 2} f(y) \, dy\right) \, dx.
$$

Proof.
We have
$$
\begin{aligned}
P(X_{(1)} + X_{(n)} \leq a)
&= \int_{-\infty}^\infty \int_{-\infty}^{a - x} g(x, y) \, dy \, dx
\end{aligned}
$$
Since $g(x, y) = 0$ if $x > y$, this integral reduces to
$$
\int_{-\infty}^{a/2} \int_x^{a - x} g(x, y) \, dy \, dx
$$
(draw the picture of the intersection of the regions $x \leq y$ and $x + y \leq a$ to see why).
Now just plug in the density formula from Proposition 1.
