# Existence of $\lim \frac{f(x)}{x}$ when $f \in C[0,+\infty)$, $f(x) \ge 0$ and $f(x+y) \le f(x) + f(y)$.

$$f(x) \in C[0,+\infty)$$, $$f(x) \ge 0$$ and $$f(x+y) \le f(x) + f(y)$$. Must $$\lim_{x \rightarrow +\infty} \frac{f(x)}{x}$$ exist?

Here is my thought. First, it's easy to see that $$\lim_{n \rightarrow +\infty} \frac{f(2^nx)}{2^nx}$$ exists for every $$x>0$$ since $$0 \le \frac{f(2^nx)}{2^nx} \le \frac{f(2^{n-1}x)+f(2^{n-1}x)}{2^nx} = \frac{2^{n-1}x}{2^{n-1}x}$$ so $$\{ \frac{f(2^nx)}{2^nx} \}$$ is monotonically decreasing and bounded. Let $$\frac{f(2^n)}{2^n} \rightarrow a$$ and if $$\lim\frac{f(x)}{x}$$ doesn't exist then there is $$x_{n}$$ such that $$x_n \rightarrow +\infty$$ and $$\frac{f(x_n)}{x_n} \ge \delta + \frac{f(2^{N_n})}{2^{N_n}}$$ where $$2^{N_n} \le x_n \le 2^{N_n+1}$$ when $$n$$ is great enough (The situation when left side $$\le$$ right side is similar).

Therefore we have $$f(x_n - 2^{N_n}) + f(2^{N_n}) \ge f(x_n) \ge \delta x_n + \frac{f(2^{N_n})}{2^{N_n}} x_n \Rightarrow \\ f(x_n - 2^{N_n}) \ge \delta x_n + \frac{f(2^{N_n})(x_n - 2^{N_n})}{2^{N_n}}$$ Since $$x_n \rightarrow +\infty$$ we have $$f(x_n - 2^{N_n}) \rightarrow +\infty$$ and so do $$\{ x_n - 2^{N_n} \}$$. And therefore $$\frac{f(x_n - 2^{N_n})}{x_n - 2^{N_n}} \ge \delta \frac{x_n}{x_n - 2^{N_n}} + \frac{f(2^{N_n})}{2^{N_n}}$$ Since $$2^{N_n} \le x_n \le 2^{N_n+1}$$ we have $$\frac{x_n}{x_n - 2^{N_n}} \ge \frac{3}{2}$$. Let $$b_n = x_n -2^{N_n}$$ so that $$b_n \rightarrow \infty$$ and $$\frac{f(b_n)}{b_n} \ge \delta \frac{3}{2} + \frac{f(2^{N_n})}{2^{N_n}}$$. And therefore $$\frac{b_n}{b_n} \ge \frac{3}{2} \delta + a$$ and $$\frac{b_n}{b_n} \ge \frac{3}{2} \delta + \frac{f(2^{N'_n})}{2^{N'_n}}$$ when $$n$$ is great enough and therefore we have $$\frac{f(c_n)}{c_n} \ge \frac{9}{4} \delta + a$$ that is for any $$N>0$$ we have $${s_n} \rightarrow \infty$$ such that $$\frac{f(s_n)}{s_n} \ge N$$.

But I haven't found any contradiction in it.

By Fekete's lemma, $$\lim \frac{f(n)}{n} = \inf_{n \ge 1}\frac{f(n)}{n}=L$$.

So $$\lim_{x \rightarrow +\infty}\frac{f(\lfloor x \rfloor)}{\lfloor x \rfloor} = \lim_{x \rightarrow +\infty}\frac{f(\lceil x \rceil)}{\lceil x \rceil} = L$$.

Observe that $$\lfloor x \rfloor \le x \le \lceil x \rceil$$ and that $$f$$ is bounded on $$[0, 1]$$.

Since

$$\frac{f(x)}{x} \le \frac{f(\lfloor x \rfloor)}{x} + \frac{f(x - \lfloor x \rfloor)}{x} \le \frac{f(\lfloor x \rfloor)}{\lfloor x \rfloor} + \frac{f(x - \lfloor x \rfloor)}{x}$$

we have $$\limsup_{x \rightarrow +\infty} \frac{f(x)}{x} \le L$$.

Analogously, from

$$\frac{f(x)}{x} \ge \frac{f(\lceil x \rceil)}{x} - \frac{f(\lceil x \rceil - x)}{x} \ge \frac{f(\lceil x \rceil)}{\lceil x \rceil} - \frac{f(\lceil x \rceil - x)}{x}$$

we get $$\liminf_{x \rightarrow +\infty} \frac{f(x)}{x} \ge L$$.

Therefore, $$\lim_{x \rightarrow +\infty} \frac{f(x)}{x} = L$$.