if $f(\frac{x+y}{2})=\frac{1}{2}[f(x)+f(y)], f(0)=0,f(1)=1$ then$f(\frac{1}{22})=?$ let function $f:[0,1]\to [0,1]$,and such $f(0)=0,f(1)=1$,
and foy any $0\le x\le y\le 1$,then we have
$$f\left(\dfrac{x+y}{2}\right)=\dfrac{1}{2}[f(x)+f(y)]$$
Question 1
Find the value $f(\dfrac{1}{22})$
Qusetion 2: Find the 
$$f(\dfrac{1}{n})=\dfrac{1}{n}?,n\in N^{+}$$
My try: Now I have solve question 1:
since
$$f\left(\dfrac{x+y}{2}\right)=\dfrac{1}{2}[f(x)+f(y)]$$
then let $x=0$,we have
$$f(y)=2f(\dfrac{y}{2})$$
let $f(\dfrac{1}{22})=a$,then
$$f(\dfrac{1}{11})=f(\dfrac{2}{22})=2f(\dfrac{1}{22})=2a,f(\dfrac{2}{11})=4f(\dfrac{1}{22})=4a,f(\dfrac{4}{11})=8f(\dfrac{1}{22})=8a$$
and $$f(\dfrac{8}{11})=16f(\dfrac{1}{22})=16a$$
and note
$$f(\dfrac{6}{11})=f(\dfrac{\dfrac{1}{11}+1}{2})=\dfrac{1}{2}(f(\dfrac{1}{11}+1)=\dfrac{1}{2}(2a+1)$$
and other hand
$$f(\dfrac{6}{11})=f(\dfrac{4}{11}+\dfrac{8}{11}/2)=\dfrac{1}{2}(8a+16a)$$
so
$$\dfrac{1}{2}(8a+16a)=\dfrac{1}{2}(2a+1)\Longrightarrow a=\dfrac{1}{22}$$
so
$$f(\dfrac{1}{22})=\dfrac{1}{22}$$
But question 2,How prove it? Thank you 

As long as you consider $f$ at rational places, no continuity assumption is needed. Since 1. is a special case of 2., let's solve that:
Let $n\in\mathbb N$ and $x_i=\frac in$, $0\le i\le n$.
Then for $0<i<n$ we have $x_i=\frac{x_{i-1}+x_{i+1}}2$ and hence $f(x_i)=\frac{f(x_{i-1})+f(x_{i+1})}2$, i.e. $$\tag1f(x_{i+1})=2f(x_i)-f(x_{i-1}).$$ Then by induction we find that $f(x_i)=if(x_1)$ for $0\le i\le n$. Indeed, this is trivially true for $i=0$ and for $i=1$ and the induction step follows immediately from $(1)$. From $nf(x_1)=f(x_n)=f(1)=1$ we then conclude that $f(x_1)=\frac1n$, as was to be shown.
(In fact, we have at the same time shown that $f(x)=x$ for all $x\in[0,1]\cap \mathbb Q$ as $\frac mn$ with $0\le m\le n$ occurs as $x_m$ in the above sequence and $f(x_m)=mf(x_1)=\frac mn$)
