Let $(u_n)_{n >= 0}$ a sequence of positive numbers, such as
1) $u_0 =1$
2) for all integer $n>0$, at least half of the $u_0, u_1, ..., u_{n-1}$ are superior or equals to $2u_n$.
Show that $u_n$ converges to $0$.
My thoughts :
If $u_0 = 1$, then
$u_1 <= \frac{u_0}{2} $. Otherwise, 2) doesn't hold.
$u_2 <= \frac{max(u_0, u_1)}{2} <= \frac{max(u_0,\frac{u_0}{2})}{2} = \frac{u_0}{4}$. Otherwise, 2) doesn't hold.
$u_3 <= \frac{max(min(u_0,u_1), u_2)}{2} = \frac{max(u_1,u_2)}{2} = \frac{u_0}{4}$. Otherwise, 2) doesn't hold.
...
$u_n <= \frac{max(u_{n-1}, u_{n-2})}{2} <= \frac{u_0}{2^n} =\frac{1}{2^n} $. Otherwise, 2) doesn't hold.
Am I right? How would I prove that recursively ?
I think that your inequality on $u_2$ is not proved, as ${\rm max}\{u_0,\frac{u_0}{2}\}=u_0=1$.
I try a solution:
For $n\geq 2$, I put $\displaystyle v_n={\rm max}\{u_k,\frac{n-1}{2}\leq k\leq n-1\}$. By the hypothesis, there exist $k$, $\displaystyle \frac{n-1}{2}\leq k\leq n-1$, such that $2u_n\leq u_k$. Hence we get $2u_n\leq v_n$.
Now $$v_{n+1}={\rm max}\{u_k,\frac{n+1}{2}\leq k\leq n\}= {\rm max}\{ {\rm max}\{u_k,\frac{n+1}{2}\leq k\leq n-1\}, u_n\}\leq {\rm max}\{v_n, \frac{v_n}{2}\}=v_n$$ Hence $v_n$ is decreasing and positive, so $v_n\to L\geq 0$. Suppose that $L>0$. Then, there exists $N$ such that for $n\geq N$, we have $\displaystyle v_n\leq \frac{3L}{2}$. Then for $n\geq N$, we have $\displaystyle u_n\leq \frac{3L}{4}$. For $\displaystyle \frac{n-1}{2}\geq N$, this gives $\displaystyle v_n\leq \frac{3L}{4}$, and $\displaystyle L\leq \frac{3L}{4}$, a contradiction. Hence $L=0$, and of course, $u_n\to 0$.