Prove a Continuous Function Bounded

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function which satisfies $$\sup_{x,y\in\mathbb{R}}|f(x+y)-f(x)-f(y)|<\infty$$ If we have $\displaystyle{\lim_{n\rightarrow\infty,n\in\mathbb{N}}\frac{f(n)}{n}=2014}$, prove $\displaystyle{\sup_{x\in\mathbb{R}}|f(x)-2014x|}<\infty$.

(Problem 1 in S.-T. Yau College Student Mathematics Contests 2014, Analysis and Differential Equations Individual)

Firstly we can replace $2014$ by $0$ so we actually need to prove $f(x)$ to be bounded over $\mathbb{R}$. And it's not difficult to prove that $\displaystyle{\lim_{x\rightarrow+\infty}\frac{f(x)}{x}=0}$, but idk how to go on with it. Thanks for your help!!!

Answer 1

WLOG, consider a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that

• $g(0)=0$,
• $g([0,+\infty)) \supset [0,+\infty)$,
• $|g(x+y)-g(x)-g(y)| \le 1$ for any $x,y\in\mathbb{R}$.

Since $g$ is continuous, we can define $$ a_k := \inf \{ x>0 : g(x)=2^k \}, \quad k=1,2,\cdots. $$ It is clear that $a_k$ increases to $+\infty$ as $k\to\infty$. Also, we will find that $$ g(a_k)/a_k \ge 1/a_1 >0, $$ which provides a contradiction.

Note that $a_1 \in (0,+\infty)$ and $g(a_1)=2$. If $a>0$ and $g(a)=n>0$, then $$ |g(a+a_1)-g(a)-g(a_1)| \le 1 \implies g(a+a_1) \ge n+1, $$ and $$ |g(2a+a_1)-g(a)-g(a+a_1)| \le 1 \implies g(2a+a_1) \ge 2n. $$ Therefore, we have $a_{k+1} \le 2 a_k + a_1$, and thus $$ a_{k+1}+a_1 \le 2(a_k+a_1) \implies a_k+a_1 \le 2^{k-1}(a_1+a_1). $$ It follows that $a_k \le 2^k a_1$ and the proof is complete.