Bounded and Divergent sequence with $x_{n+m}\leq (x_n+x_m)/2$ My question is: Does there exist $x_n$ ($n\geq 0$) such that $x_n$ is a bounded and divergent sequence with $$x_{n+m}\leq (x_n+x_m)/2$$ for all $n,m\geq 0$?
I'm guessing that such an example does exist but can't seem to find an example.
Since we require $x_n$ to be bounded, we can't have something with $\lim_{n\to\infty} x_n\to\pm \infty$, so it has to look something like $x_n=(-1)^n$, although this doesn't work since
$$(-1)^{1+1}=1\not\leq ((-1)^1+(-1)^1)/2=-1$$
Assuming that such a sequence can't be found, I'm not sure how I'd prove it either.
 A: Assume that a bounded sequence $(x_n)_{n\ge 0}$ satisfies $\displaystyle x_{n+m} \le \frac{x_n+x_m}{2}$ for arbitrary $n,m\ge 0$. Let $\displaystyle a=\liminf_{k\to\infty} x_k$ and $\displaystyle b=\limsup_{k\to\infty} x_k$. Since the sequence $(x_n)_{n\ge 0}$ is bounded, we have $-\infty <a\le b<\infty$. Applying $\displaystyle\limsup_{m\to\infty}$ to both sides of the given inequality we obtain $$b = \limsup_{m\to\infty} x_{n+m} \le \limsup_{m\to\infty} \frac{x_n+x_m}{2} = \frac 12 x_n+\frac b2$$ for every $n\ge 0$. Applying $\displaystyle \liminf_{n\to\infty}$ we arrive at $$b \le \frac 12 a+\frac 12 b$$ which gives $b\le a$. Since $a \le b$, it follows that $a=b$ and the sequence $(x_n)_{n\ge 0}$ converges.
A: Let use denote $S = \limsup_{n \to \infty} x_n $. Fix an index $n$ and choose a subsequence $(m_k)$ such that $x_{n + m_k} \to S$. Then
$$
S = \limsup_{n \to \infty} x_{n + m_k} \le \limsup_{n \to \infty} \frac 12 (x_n + x_{m_k}) \le \frac 12 (x_n + S)
$$
which implies that $S \le x_n$. This holds for all $n$, so that
$$
 S \le \liminf_{n \to \infty} x_n
$$
which means that the sequence is convergent.
If we drop the boundedness condition then the same method shows that the sequence converges to a finite value or diverges to $+\infty$ or to $-\infty$.
