The problem states to prove that for any $m, n \in \mathbb{N}$, if $0 < x_{m+n} \leq x_m + x_n$, then the sequence $\left\{\frac{x_n}{n}\right\}$ has a limit $\lim_{n \to \infty} \frac{x_n}{n}$ that exists.

Proof:

1. Step 1: Define the sequence. Let $a_n = \frac{x_n}{n}$. We aim to show that the sequence $\{a_n\}$ converges, meaning $\lim_{n \to \infty} a_n$ exists.

2. Step 2: Use the inequality. From the given inequality $0 < x_{m+n} \leq x_m + x_n$, divide through by $m+n$: $\frac{x_{m+n}}{m+n} \leq \frac{x_m}{m+n} + \frac{x_n}{m+n}.$

3. Step 3: Express bounds for $a_n$. Let $a_n = \frac{x_n}{n}$. Rewriting the inequality gives: $\frac{x_{m+n}}{m+n} \leq \frac{m}{m+n} \cdot \frac{x_m}{m} + \frac{n}{m+n} \cdot \frac{x_n}{n}.$ This can be simplified as: $a_{m+n} \leq \frac{m}{m+n} a_m + \frac{n}{m+n} a_n.$

4. Step 4: Subadditivity of the sequence. The inequality above shows that the sequence $\{a_n\}$ is subadditive, which implies that $\{a_n\}$ has a well-defined limit as $n \to \infty$ due to the Fekete subadditive lemma.

5. Step 5: Existence of the limit. Since $\{a_n\}$ is bounded below by 0 (as $x_n > 0$) and satisfies the subadditivity property, the sequence $\{a_n\}$ converges. Thus, $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{x_n}{n}$ exists.

Conclusion:

The sequence $\left\{\frac{x_n}{n}\right\}$ has a limit, and $\lim_{n \to \infty} \frac{x_n}{n}$ exists.

Let me know if you'd like a detailed explanation of any specific step.

Follow-up Questions:

1. What does the subadditivity property imply about a sequence's behavior?
2. How does dividing by $m+n$ simplify the given inequality?
3. Can this proof be extended to other types of sequences with similar inequalities?
4. How does the Fekete subadditive lemma guarantee convergence of the sequence?
5. What are some examples of subadditive sequences, and how do they behave?

Tip:

When working with subadditive or superadditive sequences, always check if the sequence is bounded. This often helps prove convergence efficiently.