Let $(y\_n)\_{n\ge 1}$ be a sequence with values in $(0,1)$ such that $\lim\_n y\_n=1$. Let also $f: [0,1]\to \mathbb{R}$ be a bounded function such that $f(0)=0$ and satisfies
$$
\forall n\ge 1, \forall x \in [0,1-y\_n], \quad 
f(x+y\_n)=f(x)+f(y\_n)
\quad \quad (\star)
$$
and
$$
\forall x \in [0,1], \quad f(x)+f(1-x)=0.
\quad \quad (\star\star)
$$



> 
> **Question.** Is it true that $f$ is constantly $0$ in a neighborhood of $0$?
> 
> 
> 


Note: let $\mathscr{B}$ be a basis of $\mathbb{R}$ over $\mathbb{Q}$ such that $1 \in \mathscr{B}$. Pick $b \in \mathscr{B}$ with $b\neq 1$ and, for each $x \in \mathbb{R}$, let $h(x)$ be the (unique rational) coefficient of $x$ with respect to $b$ in the representation of $x$ as finite linear combination of elements of $\mathscr{B}$. Let $f$ be the restriction of $h$ to $[0,1]$. Then $f(0)=f(1)=0$ and $f(x+y)=f(x)+f(y)$ for all $x,y \in [0,1]$ with $x+y \in [0,1]$. In addition, $f(1-x)=h(1)+h(-x)=-h(x)=-f(x)$. Hence $(\star)$ and $(\star \star)$ hold. However, it is known that the graph of $h$ is dense in $\mathbb{R}^2$, hence $f$ is *not* bounded.



Symmetric and nearly additive bounded functions
Let $C$ be the usual Cantor set in $[0,1]$, and split the complement $I\setminus C$ as a disjoint union of intervals $I\_1,I\_2,\dots$. Let $L\_i$ and $R\_i$ denote the left and right half of $I\_i$, and let $m\_i$ be the midpoint of $I\_i$.


Then the function
$$
 f(x)=
 \begin{cases}
 0, & x\in C;\\
 1, &x\in L\_i; \\
 -1, & x\in R\_i; \\
 0, & x=m\_i
 \end{cases}
$$
satisfies the requirements with $y\_i=1-3^{-i}$, though it does not vanish in any neoghbourhood of $0$.


With a bit of care, $f$ can be made continuous.


