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$f(x) \in C[0,+\infty)$, $f(x) \ge 0$ 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.

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1 Answer 1

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

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