Potla wrote: Suppose $f:\mathbb{R} \to \mathbb{R}$ is a function such that \[|f(x+y)-f(x)-f(y)|\le 1\ \ \ \text{for all} \ \ x, y \in\mathbb R.\] Prove that there is a function $g:\mathbb{R}\to\mathbb{R}$ such that $|f(x)-g(x)|\le 1$ and $g(x+y)=g(x)+g(y)$ for all $x,y \in\mathbb R.$ Let $x\in\mathbb R$ and the sequence $a_n=2^{-n}f(2^nx)$ We get $|f(2^nx+2^nx)-2f(2^nx)|\le 1$ and so $|a_{n+1}-a_n|\le 2^{-n-1}$ So, considering $n>p$ : $|a_n-a_p|\le 2^{-n-1}+2^{-n}+...+2^{-p-1}<2^{-p}$ And so $a_n$ is a Cauchy sequence and so is convergent. Let then $g(x)=\lim_{n\to +\infty}2^{-n}f(2^nx)$ $|f(2^n(x+y))-f(2^nx)-f(2^ny)|\le 1$ and so $|2^{-n}f(2^n(x+y))-2^{-n}f(2^nx)-2^{-n}f(2^ny)|\le 2^{-n}$ Setting $n\to+\infty$ in this inequality, we get $g(x+y)=g(x)+g(y)$ From $|a_n-a_p|<2^{-p}$, we also get $|a_n-a_0|<1$ and so $|a_n-f(x)|<1$ Setting $n\to+\infty$ in this inequality, we get $|g(x)-f(x)|\le 1$ Q.E.D.

Exceptional problem. Solved with Jeffrey Chen, Luke Robitaille, and Raymond Feng. Evidently we have \[|f(nx)-nf(x)|\le n-1\tag{\(\star\)}\]for all positive integers \(n\). Claim: The limit \[\lim_{n\to\infty}\frac{f(nx)}n\]exists. Proof. We have by \((\star)\) that \begin{align*} |f(mnx)-mf(nx)|&\le m-1\\ |f(mnx)-nf(mx)|&\le n-1. \end{align*}It follows that \[\left\lvert\frac{f(mx)}m-\frac{f(nx)}n\right\rvert\le\frac{m+n-2}{mn},\]implying the sequence \(\frac{f(x)}x\), \(\frac{f(2x)}{2x}\), \(\frac{f(3x)}{3x}\), \(\ldots\) is Cauchy, so it converges by completeness of \(\mathbb R\). \(\blacksquare\) Now we claim the function \[g(x):=\lim_{n\to\infty}\frac{f(nx)}n\]works. We have \[ \left\lvert f(x)-g(x)\right\rvert =\lim_{n\to\infty}\frac{|nf(x)-f(nx)|}n \le\lim_{n\to\infty}\frac{n-1}n=1 \]by $(\star)$. We have \[ g(x+y)-g(x)-g(y) =\lim_{n\to\infty}\frac{f(n(x+y))-f(nx)-f(ny)}n=0 \]by the squeeze principle, since \[-\frac1n\le\frac{f(n(x+y))-f(nx)-f(ny)}n\le\frac1n.\] This completes the proof. Remark: Over \(\mathbb Q\), the problem is false by taking \[f(x)=\lfloor\pi x\rfloor.\]

Potla wrote: Suppose $f:\mathbb{R} \to \mathbb{R}$ is a function such that \[|f(x+y)-f(x)-f(y)|\le 1\ \ \ \text{for all} \ \ x, y \in\mathbb R.\]Prove that there is a function $g:\mathbb{R}\to\mathbb{R}$ such that $|f(x)-g(x)|\le 1$ and $g(x+y)=g(x)+g(y)$ for all $x,y \in\mathbb R.$ Cute. Define \(a_m=f(mx)\) for a fixed real \(x\). We clearly have \[\lvert a_{mn}-na_m\rvert\leq n-1\]Define \(g(x)\) to be the limit of \(\frac{a_m}{m}\). Note that \(\left\{\frac{a_m}{m}\right\}\) is the Cauchy sequence, because \[\left\lvert\frac{a_m}{m}-\frac{a_n}{n}\right\rvert\leq\frac{1}{m}+\frac{1}{n}\]and so it converges. Also \[g(x+y)-g(x)-g(y)=\lim_{n\rightarrow\infty}\frac{f(n(x+y))-f(nx)-f(ny)}{n}=0\]so \(g\) is additive. Finally, \[\lvert f(x)-g(x)\rvert=\lim_{n\rightarrow\infty}\frac{1}{n}\lvert a_n-na_1\rvert\leq\frac{n-1}{n}\leq1\]as desired. So \(g\) is our wanted function.

EDIT: My construction is actually equivalent to theULTIMATE geo god's construction, so it's correct Let $x$ be a fixed positive real. Let $a_j=f(jx)$. Claim: There exists a unique positive real, call $c$, such that $|a_j-cj|\le 1$ for all $j\in \mathbb{Z}$. Proof: If it exists, its uniqueness is clear because if $c,d$ both work, then $|a_j-cj| + |a_j-dj| \le 2$ but$|a_j-cj| + |a_j-dj| \ge (d-c)j$. This is contradiction when $j$ is sufficiently large.! Now I prove it exists. Let $X$ be the set of reals $c$ such that there exists $j$ such that $a_j-cj\le -1$ and $Y$ be the set of reals $c$ such that there exists $j$ such that $a_j-cj\ge 1$. Note $X\cap Y=\emptyset$ is necessary because all elements in $X$ are too large to be $c$ and all elements in $Y$ are too small to be $c$. It is also sufficient because $X,Y$ are open intervals. Suppose $c\in X$ and $c\in Y$. Take $a_j\le cj-1$ and $a_k\ge ck+1$. Then $a_{jk}\ge ja_k-(j-1)\ge cjk+1$ by induction. Similarly, $a_{jk}\le ka_j+(j-1)\le cjk-1$. These bounds contradict each other, so I am done. I can take $c=\mathbb{R}\backslash (X\cup Y)$ Now, let $c_x$ be the unique real such that $|f(jx)-c_xj|\le 1$ for all $j\in \mathbb{Z}$. We can prove that the inequality holds for all $x\in \mathbb{Q}$. Note $c_{x+y}=c_x+c_y$ because otherwise we have $1\ge |f(xj+yj)-f(xj)-f(yj)|\ge |c_{x+y}-c_x-c_y|j - 2$, which is false for $j$ large. Therefore, we let $g(x)=c_x$. We know $g$ is additive, and $|f(x)-c_x|\le 1$ by definition. The conclusion follows.

The same problem was given at a Bulgarian TST for IMO 2020 (p5). You can see a different approach in this link. There was an extra statement that $g$ must be unique, which is the easiest part.