Let $d\geq 1$ be a positive integer. If $\{\vec x_n\}_{n=1}^\infty$ is a sequence of $d$-dimensional vectors satisfying $$\lvert\vec x_{n+m}\rvert\leq \lvert\vec x_n+\vec x_m\rvert\qquad \text{for all }n,m\in\mathbb N_{\geq 1},\tag{$*$}\label{star}$$
where $\lvert\vec x\rvert=[(x^{(1)})^2+\dotsb+(x^{(d)})^2]^{1/2}$ for $\vec x=(x^{(1)}, \dotsb, x^{(d)})$. Then does the limit $\lim\limits_{n\to\infty}\frac{\vec x_n}{n}$ exist?
This has been asked in MSE, without getting any convincing answer.
Ideas. It follows from \eqref{star} and the induction that $\lvert\vec x_n\rvert\leq n\lvert\vec x_1\rvert$ for all $n\in\mathbb N_{\geq 1}$, hence $\{\vec x_n/n\}_{n\geq 1}$ is a bounded sequence. On the other hand, \eqref{star} also implies that $\lvert\vec x_{n+m}\rvert\le \lvert\vec x_n\rvert+\lvert\vec x_m\rvert$ for all $n,m\in\mathbb N_{\geq 1}$, hence by Fekete's subadditive lemma we get the convergence of $\{\lvert\vec x_n\rvert/n\}_{n\geq 1}$ with
$$\lim_{n\to\infty}\frac{\lvert\vec x_n\rvert}n=\inf_{n\geq 1}\frac{\lvert\vec x_n\rvert}n.$$
If $d=1$, then the answer is yes, which is proved in Display name's answer to Does $|x_{n+m}|\leq |x_n+x_m|$ imply the convergence of $\{x_n/n\}$?. For $d\geq2$, I tried to use the same ideas as in that post. If $\lim\limits_{n\to\infty}\frac{\lvert\vec x_n\rvert}n=L$, then for all $\epsilon>0$ there exists $N\gg L$ such that $\left\lvert\frac{\lvert\vec x_n\rvert}n-L\right\rvert<\varepsilon$ for all $n>N$. Take $\varepsilon>0$ small enough. Fix $n_0>N$, then there exists $\vec z_1$ with $\lvert\vec z_1\rvert=1$ such that $\left\lvert\frac{\vec x_{n_0}}{n_0}-L\vec z_1\right\rvert<\varepsilon$. Let $n$ be the largest integer such that $\left\lvert\frac{\vec x_{k}}{k}-L\vec z_1\right\rvert<\varepsilon$ holds for all $n_0\leq k\leq n$. Then $\left\lvert\frac{\vec x_{n}}{n}-L\vec z_1\right\rvert<\varepsilon$ and $\left\lvert\frac{\vec x_{n+1}}{n+1}-L\vec z_1\right\rvert\geq\varepsilon$, hence $\left\lvert\frac{\vec x_{n+1}}{n+1}-L\vec z_2\right\rvert\geq\varepsilon$ for some $\vec z_2$ with $\lvert\vec z_2\rvert=1$ and $\vec z_2$ is far away from $\vec z_1$. However, they are not as far away as in that post, where $\vec z_1=1$, $\vec z_2=-1$ for $d=1$. I'm not sure how to continue.
Best Answer
The answer is still (surprisingly to me!) yes. As you've already established, there must exist $L = \lim \frac{||x_n||}{n}$ (I will never write a vector symbol over the variable). If $L = 0$ then there is nothing to prove. Otherwise, assume for simplicity that $L = 1$ by scaling. Let $x_n = c_n v_n$, $||v_n|| = 1$, $\lim \frac{c_n}{n} = 1$.
First, obeservation: for all $s > 1$ and all $\delta > 0$ there exists $N$ such that for all $N \le n \le m \le sn$ we have $\langle v_n, v_m \rangle \ge 1 - \delta$ (that is, the angle between $v_n$ and $v_m$ is small). Indeed, just go far enough in the sequence so $\frac{c_n}{n}$ is very close to $1$, then if the angle is big enough, by the law of cosines (and comparable sizes of $c_n$ and $c_m$ due to the $s$ restriction) we will get that $c_{n+m} < (n+m)$ -- a contradiction.
Now, our goal is to prove that the sequence of $v_n$ is Cauchy, then its limit will be our limit as well (in particular, I believe that the compactness of the unit sphere is not needed and so this proof will work even in the Hilbert space).
So, assume that we picked some very large number $N$ and nevertheless for some $N < n < m$ we have $\langle v_n, v_m \rangle < 1 - \delta_0$ (that is, the angle is not small). Importantly, $\delta_0 > 0$ is fixed but $N$ we can get as big as we like later. Then, from the above observation we know that $m$ is much bigger than $n$. Direct computation with law of cosines (and the fact that $c_m$ is much bigger than $c_n$) then shows that $||x_n + x_m|| \le c_m + \gamma c_n$ for some $\gamma < 1$ (this $\gamma$ is related to the $\delta_0$ above, but making it explicit is not pleasant). So, $c_{m+n} \le c_m + \gamma c_n$.
Here is the idea: $n+m$ and $m$ are still very close since $n$ is much smaller than $m$. So, we can invoke the above observation telling us that the angle between $v_{m+n}$ and $v_m$ is small. But this means that the angle between $v_{m+n}$ and $v_n$ is not small. So, we can repeat this procedure again and get that $c_{m+2n} \le c_m + 2\gamma c_n$.
How many times can we repeat this? By picking $N$ big enough, at least until $m + kn \le sm$. Picking the biggest such $k$ we get $c_{m+kn} \le c_m + k\gamma c_n$. Recall that $c_{m+kn} \ge m + kn$. So, combining this we get $$m + kn \le c_m + k\gamma c_n.$$
Finally, if $N$ is big enough $c_n < (1+\varepsilon)n$, $c_m < (1+\varepsilon)m$. Plugging this in we get $$m+kn \le (1+\varepsilon)(m + k\gamma n).$$
Recall that $kn = m(s-1)$ (the right hand side might not be divisible by $n$, but adjusting it is not a big deal since once again $n$ is much smaller than $m$). So,
$$sm \le (1+\varepsilon)(m + \gamma m(s-1)).$$
Dividing by $m$, we get $s \le (1+\varepsilon)(1 + \gamma (s-1))$, which is equivalent to (if $(1+\varepsilon)\gamma < 1$, which we can always achieve taking $\varepsilon$ small enough) $$s < \frac{(1+\varepsilon)(1-\gamma)}{1-(1+\varepsilon)\gamma}.$$
But the observation works for any $s > 1$ -- contradiction! (on further thought, by picking $\varepsilon$ arbitrarily small it is enough to use the observation for any fixed $s_0 > 1$, but since this version is clearly equivalent to the version with arbitrary $s > 1$ I don't think it is any easier, but also makes the ideas less obvious).