Find all functions $f:[0;1]\rightarrow[0;1]$ which satisfying the following conditions: $f(0)=0, f(1)=1$ $$f(x+y)-f(x)=f(x)-f(x-y)$$ for any $x,y\ge0$ such that $(x-y), (x+y) \in [0;1]$.
Let $x=y$. Then $$f(2x)=2f(x)$$ for $0\le x \le \frac12$
Let $y=2x, y=3x,...$. Then $$f(nx)=nf(x)$$ for $0\le x \le \frac1n$
Edit: This is an IMO1986 Longlisted problem (FIN 1)
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Picking up essential points from @wythagoras: For $x,y,x+y\in [0,1]$ the functional equation gives $$ 2f \left( \frac{x+y}{2} \right) = f(x+y) + f(0) = f(x)+f(y) $$ and $f(0)=0$ shows that $$ f(x+y)=f(x)+f(y)$$ $f\geq 0$ implies that $f$ is monotone increasing. Iterating we get $1=f(1)=n f(1/n)$ and then $f(k/n)=k/n$ for $0\leq k\leq n$, $n\geq 1$. Then for any real $x\in (0,1)$ (and rational q's): $$ x=\sup\{q<x\} \leq f(x) \leq \inf\{q>x\} =x$$ The important point is positivity (or boundedness) of $f$. Without this it is possible to construct other (non-measurable) functions using a Hamel basis of the reals over ${\Bbb Q}$. See: Cauchy's Functional equation
From $f(nx)=nf(x)$ you get $1=f(1)=n f(\frac1n)$, so $f(\frac1n)=\frac1n$. Then $f(\frac{m}{n})=m f(\frac1n)=\frac{m}{n}$ for all $0<m<n$. So $f(x)=x$ for all rational $0<x<1$.
This gives $f(x)=x$ for all real $0\leq x \leq 1$ if $f$ is strictly increasing.
To prove $f$ is (non-strictly) increasing (i.e. $a>b \implies f(a) \geq f(b)$, ), we substitute $z=x+y$, $t=x-y$ and assume $z+t\leq1$. Then $f(z)+f(t)=2f(\frac12(z+t))=f(z+t)$, so if $a>b$ then let $z=b$ and $t=a-b$, then $$f(a)=f(b)+f(a-b) \geq f(b)$$
Now note that $f(c)=0$ for some $c>0$, then $f(d)=0$ for all $0<d<c$. Take any reciprocal of an integer $\frac1q$ such that $0<q<c$, then and by using $f(nx)=nf(x)$ we get $f(1)=f(q \frac1q)=q f(\frac1q)=0$, contradiction. So we get in fact that $f(x)>0$ for $x>0$, so $$f(a)=f(b)+f(a-b) > f(b)$$
And hence is $f$ strictly increasing.