Function and sets 
Consider the functions $f:N_0 \to {N_0}$ ,where $N_0$ is the set of all non negative integers, defined by the following conditions
(i) $f(0)=0$ (ii) $f(2n)=2f(n)$ (iii) $f(2n+1)=n+2f(n)$
Determine the following three sets
1)$L\equiv \text{{n|f(n)>f(n+1) }}$
2)$E \equiv \text {{n|f(n)=f(n+1)}}$
3)$G\equiv \text{{n|f(n)<f(n+1)}}$
Also find for each $k\geq 0$ find formula for $a_k \equiv \text{{ max f(n);$0\leq n \leq 2k$}}$ in terms of $k$

Now I have found one subset each of  $L ,E$ as follows
For $L_1$ we have
for any $k \geq 0$ $f(2k)-f(2k+1)=-k<0$ ,
and  For $E_1$ we have $f(4k+1)=2f(2k)+2k=4f(k)+2k$
and $f(4k+2)=2k+4f(k)$
But for the subset $G_1$ of $g$ I cannot seem to make a dent in the problem. Also, I am unsure if there are other subsets of $L$ and $E$ .
Any help is welcome
 A: You basically want to look at the sign of the forwards difference of $f$ which is $\Delta f(n) = f(n+1)-f(n)$.
As you have correctly shown, that for an even number $n$, $\Delta f(n) = n/2$ meaning $0$ belongs to $E$ and the other even numbers belong to $G$.
For an odd number $n$ there are 2 cases: if $k$, such that $n=2k+1$, is even you have correctly shown that $n\in E$. If $k$ is odd, you can show by strong induction that the forwards difference is negative at such odd numbers $n$. Meaning $n\in L$.
Since $L, E, G$ are a partition of $N_0$ and so are the nonzero even numbers, let's call them $2N$, the odd numbers where the quotient $\mod 2$ is even, let's call them $O_1$ (note, we are also including $0$ in here) and the odd numbers where the quotient $\mod 2$ is odd, let's call them $O_2$, and you have shown that $2N\subseteq G,O_1\subseteq E,O_2\subseteq L$, we can conclude that $2N= G,O_1= E,O_2= L$. (Think about it, if one partition lies within another exactly matching each cell, what else could be the case?)
