Prove that $\frac{x_n}{n}$ is convergent if $x_{m+n} \ge x_m + x_n$

Question

Given a positive sequence $x_1 , x_2 , ...$ such that $x_{m+n} \ge x_m + x_n \forall m,n \in N$

Prove that $\frac{x_n}{n} \to l $ where $l$ may be a number or infinity.

Here's my original approach:

It is easy to prove that $\displaystyle \frac{x_1}{1} \le \frac{x_2}{2} \le \frac{x_4}{4} \le ...$ so the sequence $\frac{x_{2^n}}{2^n}$ has a limit c (*) (i'm supposing $\frac{x_n}{n}$ is bounded). So it is sufficient to prove that for any fixed odd number d, the sequence $\frac{x_{d.2^n}}{d.2^n}$ has a limit c too.

Any number d can be represented as $\displaystyle d=2^{e_1} + 2^{e_2} + ... + 2^{e_k}$ So $\frac{x_{d.2^n}}{d.2^n} \ge \frac{x_{2^{e_1 + n}} + x_{2^{e_2 + n}} + ... + x_{2^{e_k + n}}}{d.2^n}$ It is easy to check that the right side of this ineq tends to c using (*).

Similarly, d can also be represented as $\displaystyle d+2^{f_1} + 2^{f_2} + ... + 2^{f_t} = 2^g$ so $\frac{x_{d.2^n}}{d.2^n} \le \frac{x_{2^{g+n}}-x_{2^{f_1+n}} - x_{2^{f_2+n}} - ... - x_{2^{f_t+n}}}{d.2^n} $. Again, from (*) one can check that the right side tends to c when n tends to infinity.

Thus the proof is complete. This proof looks suspicious, so I would be glad if someone can help me verify it. Thank you.

EDIT: The proof is wrong, as pointed out in one of the answer below.

Answer 1

See here. Fekete's superadditive lemma says that $$\lim_{n\to\infty}\frac{x_n}{n}=\sup_{n\in\mathbb N}\frac{x_n}{n}.$$

Answer 2

An answer dealing with the first version of the question is now accepted hence the OP might or might not still be looking for an answer to the second version of their post, which asks to check a proof. Anyway, here is a take on that question: is the proposed proof correct?

It is easy to prove that $\frac{x_1}{1} \le \frac{x_2}{2} \le \frac{x_4}{4} \le ...$ so the sequence $\frac{x_{2^n}}{2^n}$ has a limit $c$ (*) (i'm supposing $\frac{x_n}{n}$ is bounded).

One should not have to assume that the sequence of general term $x_n/n$ is bounded, or, at least, one should treat this case and then the case when it is unbounded. Anyway, for now, let us assume that this sequence is bounded.

So it is sufficient to prove that for any fixed odd number $d$, the sequence $\frac{x_{d.2^n}}{d.2^n}$ has a limit $c$ too.

Sorry? This would be sufficient if the claim below were true.

Claim: Let $(a(n))_n$ denote a bounded sequence such that, for every odd $k$, the sequence $(a(k2^n))_n$ converges to the same limit $c$. Then the whole sequence $(a(n))_n$ converges to $c$.

The claim is not true. To see why, assume that, for every $n$ and every odd $k$, $a(k2^n)=c'$ if $n\leqslant k$ and $a(k2^n)=c$ if $n\gt k$. Then the sequence $(a(n))_n$ fits the hypotheses of the claim and $a(k2^k)=c'$ for every odd $k$. Since the set $\{k2^k\mid k\ \text{odd}\}$ is unbounded, if $c\ne c'$, the sequence $(a(n))_n$ does not converge to $c$ (and in fact this sequence diverges).

Hence at least this step of the proof needs to be modified.

Edit: The negative result above has a, slightly loose but perhaps more familiar, analogue, which goes as follows. Consider the doubly-indexed sequence $(x(k,n))_{k,n}$ defined by $x(n,k)=1$ if $n\lt k$ and by $x(n,k)=0$ if $n\geqslant k$. Then $\lim\limits_{n\to\infty}x(n,k)=0$ for each $k$ while $\lim\limits_{k\to\infty}x(n,k)=1$ for each $n$.

Answer 3

The sequence of ratios $r_n = \frac{x_n}{n}$ has its growth limited only from below (considering any term compared to the earlier ones, and building the sequence step by step), so it could certainly go to +infinity.

The case of interest is what happens when $r_n$ is bounded above. Let $L$ and $U$ be the lim inf and lim sup of the ratios as $n$ goes to infinity. Translated to the ratios, the inequality on the sequence is $r_{m+n} \geq \frac{mr_m + nr_n}{m+n}$. Letting $m=kp$ and $0 \leq n < p$ for large $k$ (going to infinity) and $p$ constant, this means that $r_p$ is a lower bound on $L$ for any $p$. That implies that $L \geq U$ and therefore $L=U$.

The argument did not need a separate case for $U = +\infty$, but that was not obvious before writing it.

Answer 4

Personally, it is geometrically clear yet I am not sure how to write it algebraically. Consider the real-valued function $f(x)$ with the property $f(x+y)\geqslant f(x)+f(y)$. Then $f(x)$ is concave up and with the lower bound $y=f(1)x$, and $\frac{f(a)}{a}$ is non-decreasing.