# If $f$ is a continuous function such that $|f(x+y)-f(x)-f(y)|$ is bounded and $f(n)=o(n)$, then $f$ is bounded

Suppose that $f:\mathbf{R}\to\mathbf{R}$ is a continuous function such that $$\sup_{x,y\in\mathbf{R}}|f(x+y)-f(x)-f(y)|<\infty \quad (*)$$ and $\lim_{n\to\infty}\frac{f(n)}{n}=0$, prove that $\sup_{x\in\mathbf{R}} |f(x)|<\infty$

I don't know how to start, what can I get from $(*)$ ?

• I suppose that $n$ in $\lim_{n\to\infty}\frac{f(n)}{n}=0$ is an integer; this is worth emphasizing.
– user147263
Apr 4, 2015 at 19:04
• @pizza Hmm, in that case my solution is incomplete... Apr 4, 2015 at 19:38
• @pizza Ok, I think it is better now Apr 5, 2015 at 1:40

Let's try this: Set $x=y$ and get $|f(2x)-2f(x)|<M$ for some $M$ and all $x$.

Now assume that there is actually a sequence $x_n$ such that $f(x_n)\rightarrow \infty$. Find now an index $k$ such that $f(x_k)=N$ with $N$ being much greater than $M$ (how much you should decide by the rest).

Now, (*) gives us that $f(2x_k)\geq 2N-M$, then $f(4x_k)\geq 4N-2M-M$ and generally $f(2^nx_k)\geq 2^nN-a_nM$. Here $a_n$ satisfies the following:$$a_1=1,\quad a_{n+1}=2a_n+1$$ Now these imply easily that $a_n\leq 2^{n+1}-1$.

Therefore, we have $$f(2^nx_k)\geq 2^nN-2^{n+1}M$$ Now, use this to show that when $N$ is significantly greater than $M$ (you can make that rigorous), we have $\displaystyle \lim_{n\rightarrow \infty}\dfrac{f(2^nx_k)}{2^nx_k}\cong\dfrac{N}{x_k}$ which gives a contradiction.

Edit: After @pizza's comment, I'll update the proof, since the limit is assumed to be $0$ only for integers. So what we've shown by now is that the function $f$ is bounted on integers. Say that $|f(n)|\leq N$ for some $N\gg M$.

Now, $|f(n)-2f(n/2)|<M$ implies that $$M+N\geq 2|f(n/2)|\Rightarrow N\geq |f(n/2)|$$ Similarly we can show that $|f(\dfrac{n}{2^k})|\leq N$ for all $k$.

Now, just notice that the se $\{\dfrac{n}{2^k}: n,k\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ and complete the proof...

• Doesn't your recurrence relation for the $a_n$ imply that $a_n = 2^{n}-1$ exactly? Apr 4, 2015 at 18:31
• @RolfHoyer Yeah, you are right.. I wasn't paying much attention to $M$, my point was that it will be small enough to be negligible... Apr 4, 2015 at 18:34
• I think we can prove it without using proof by contradiction. Set $x=y$ we get $|f(2x)-f(x)|\leq M$, replacing $x$ by $2x$, we get $|f(4x)-2f(2x)|\leq M$, thus, $|f(4x)-f(x)|\leq 3M$. Now induction on $n$, we get $$|f(2^nx)-2^nf(x)|\leq (2^n-1)M,$$ which implies $$\left|\frac{f(2^n x)}{2^n x}-\frac{f(x)}{x}\right|\leq \frac{M}{x}$$ Let $x=\frac{m}{2^k}(m,k\in\mathbf{N})$ and send $n$ to $\infty$ yield $|f(x)|\leq M$. (Sorry for my poor English) Apr 5, 2015 at 6:10
• Yeah, this looks good! Apr 5, 2015 at 6:56

Set $L=\sup_{x,y\in\mathbf{R}}|f(x+y)-f(x)-f(y)|$. Note that \begin{align} L\geq & |f(u+v)-f(u)-f(v)|,\quad \forall u,v\in\mathbb{R}\\ L\geq & |f(r+s)-f(r)-f(s)|,\quad \forall r,s\in\mathbb{R}\\ \end{align} Due to the triangle inequality \begin{align} 2L\geq & |f(u+v)-f(u)-f(v)+f(r+s)-f(r)-f(s)|\\ \end{align} Now making $u+v=0$ and $r=0$ we have \begin{align} 2L\geq & |f(0)-f(u)-f(v)+f(0+s)-f(0)-f(s)|=|f(u)+f(v)| \end{align}. Suppose there is a sequence $x_n$ such that $\lim_{n\to \infty}|f(x_n)|=\infty$.

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Case 1: $\lim_{n\to \infty}f(-x_n)=M$. $$2L\geq |f(x_n)-f(-x_n)|$$ And so we have the inequality, $$\frac{2L}{|f(x_n)|+|f(-x_n)|}\geq \frac{|f(x_n)-f(-x_n)|}{|f(x_n)|+|f(-x_n)|} \geq \frac{|f(x_n)|}{|f(x_n)|+|f(-x_n)|} - \frac{|f(-x_n)|}{|f(x_n)|+|f(-x_n)|}$$ Making $n\to \infty$ in this inequality we have $$\frac{2L}{|f(x_n)|+|f(-x_n)|}\to 0, \mbox{ and } \frac{|f(x_n)|}{|f(x_n)|+|f(-x_n)|} - \frac{|f(-x_n)|}{|f(x_n)|+|f(-x_n)|}. \to 1$$ A contradiction.

Case 2: $\lim_{n\to \infty}|f(-x_n)|=\infty$ and $\lim_{n\to \infty}|f(x_n)|=\infty$.

$$L\geq |f(x_n-x_n)-f(x_n)-f(-x_n)|\geq |-|f(-x_n)|-|f(x_n)|+|f(0)||$$ Make $n\to \infty$ we have another contradiction.

Soon we can only conclude that there is no sequence $x_n$ such that $\lim_{n\to\infty}f(x_n)=\infty$.Therefore, $$\sup_{x\in\mathbb{R}}|f(x)|<\infty.$$