If $(x_n)_{n\in\mathbb N}$ is subadditive, show that $x_n\le\left\lfloor\frac nk\right\rfloor x_k+x_{n-k\left\lfloor\frac nk\right\rfloor}$ Let $(x_n)_{n\in\mathbb N}\subseteq[-\infty,\infty)$ with $$x_{m+n}\le x_m+x_n\;\;\;\text{for all }m,n\in\mathbb N.\tag1$$

How can we show that $$x_n\le\left\lfloor\frac nk\right\rfloor x_k+x_{n-k\left\lfloor\frac nk\right\rfloor}\;\;\;\text{for all }k,n\in\mathbb N\tag2$$ and how can we conclude that $$\limsup_{n\to\infty}\frac{x_n}n\le\frac{x_k}k\;\;\;\text{for all }k\in\mathbb N?\tag3$$

Clearly, if $$\operatorname{frac}(x):=x-\lfloor x\rfloor\in[0,1)\;\;\;\text{for }x\in\mathbb R$$ and $k,n\in\mathbb N$, we may write $$n=k\left(\left\lfloor\frac nk\right\rfloor+\operatorname{frac}\left(\frac nk\right)\right),\tag3$$ but since $\operatorname{frac}\left(\frac nk\right)$ does not necessarily belong to $\mathbb N$, I don't see how I could use the subadditivity $(1)$.
 A: As you stated, you have
$$x_{m+n}\le x_m+x_n\;\;\;\text{for all }m,n\in\mathbb{N} \tag{1}\label{eq1A}$$
and you want to show
$$x_n \le \left\lfloor\frac nk\right\rfloor x_k+x_{n-k\left\lfloor\frac nk\right\rfloor}\;\;\;\text{for all }k,n\in\mathbb{N} \tag2\label{eq2A}$$
Note \eqref{eq1A} means for all $k \in \mathbb{N}$ you have
$$\begin{equation}\begin{aligned}
x_{k+k} & \le x_k + x_k \\
x_{2k} & \le 2x_k
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Also,
$$\begin{equation}\begin{aligned}
x_{2k+k} & \le x_{2k} + x_k \\
x_{3k} & \le 2x_k + x_k = 3x_k
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
You can show quite easily, say by induction, which I'll leave to you, that for any integer $j \ge 1$ you have
$$x_{jk} \le jx_{k} \tag{5}\label{eq5A}$$
With \eqref{eq2A}, if $k \gt n$, then $\lfloor \frac{n}{k} \rfloor = 0$, with the RHS side becoming $0(x_k) + x_{n-k(0)} = x_n$, so it's quite clear \eqref{eq2A} is true. Otherwise, for $k \le n$, using \eqref{eq5A} since $\lfloor \frac{n}{k} \rfloor \ge 1$, from \eqref{eq1A} you have
$$\begin{equation}\begin{aligned}
x_{k\left\lfloor\frac{n}{k}\right\rfloor + \left(n - k\left\lfloor\frac{n}{k}\right\rfloor\right)} & \le x_{k\left\lfloor\frac{n}{k}\right\rfloor} + x_{n-k\left\lfloor\frac nk\right\rfloor} \\
x_{n} & \le \left\lfloor\frac{n}{k}\right\rfloor x_k+x_{n-k\left\lfloor\frac nk\right\rfloor}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
which shows that \eqref{eq2A} also holds for these cases.
For the other part of the question, i.e., concluding that
$$\limsup_{n\to\infty}\frac{x_n}x\le\frac{x_k}k\;\;\;\text{for all }k\in\mathbb{N }\tag{7}\label{eq7A}$$
note that when $k \mid n$, you have $\left\lfloor\frac{n}{k}\right\rfloor = \frac{n}{k}$, so \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
x_n & \le \left(\frac{n}{k}\right)x_k + x_{n - k\left(\frac{n}{k}\right)} \\
& = \left(\frac{n}{k}\right)x_k + x_{0}
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
Although $\mathbb{N}$ often doesn't include $0$, but if it's to be included for this question's purpose (although this means you also need to specify that $k \gt 0$ for equations like \eqref{eq2A}, \eqref{eq6A} and \eqref{eq8A} to make sense), note you have from \eqref{eq1A} that
$$x_{0+0} \le x_0 + x_0 \implies x_{0} \le 2x_0 \implies x_{0} \ge 0 \tag{9}\label{eq9A}$$
Alternatively, you can just assign that $x_{0} \ge 0$. Either way, this means \eqref{eq8A} becomes
$$x_n \le \left(\frac{n}{k}\right)x_k \implies \frac{x_n}{n} \le \frac{x_k}{k} \tag{10}\label{eq10A}$$
I'll leave the rest for you to finish yourself.
