Given a function$f: \mathbb{R} \mapsto \mathbb{R}$ with $f(1)=1$ and for an arbitrary $x \in \mathbb{R}$ satisfies $f(x+5) \geq f(x) + 5$ and $f(x+1) \leq f(x) + 1$. If $g(x) = f(x) - x + 1$, then evaluate the value of $g(2012)$.
My attempt: Claim: $f(x+1) = f(x)+1$
Assume that $f(x+1) < f(x)+1$ Re-arranging the assumption, we get $f(x+1)-f(x) < 1$ Now, assign values to the differences of functions such that $f(x+1)-f(x) = a$, $f(x+2)-f(x+1) = b$, $f(x+3)-f(x+2) = c$, $f(x+4)-f(x+3) = d$ and $f(x+5)-f(x+4)=e$. Adding these up, we get $a+b+c+d+e = f(x+5)-f(x)$ By our assumption, $a,b,c,d,e < 1 \Longrightarrow a+b+c+d+e < 5 \Longrightarrow f(x-5)-f(x)<5$ But, according to the given condition, $f(x+5)-f(x) \ge 5$ This contradiction occurs from our first assumption that $f(x+1) < f(x)+1$
This proves that $f(x+1)=f(x)+1$
Lemma 1: $f(x+1) = f(x)+1$
Claim: $f(x+n)=f(x)+n$ where $n \in \mathbb{Z}$ and $n \ge 0$ Assume $f(x+k) = f(x)+k$ is true Prove $f(x+k+1) = f(x)+k+1$
$f(x+k+1) = f(x+k)+1$ by lemma 1 Using our assumption, we get $f(x+k+1) = f(x+k)+1 = f(x)+k+1$
This completes the induction proving that $f(x+n) = f(x)+n$
Lemma 2: $f(x+n) = f(x)+n$ where $n \in \mathbb{Z}$ and $n \ge 0$
By the given, $g(x) = f(x)-x+1$ Substituting $x = 2012$, we get $g(2012) = f(2012)-2011$
By lemma 2, $f(2012) = f(1+2011) = f(1)+2011 = 2012$.
Substituting this back into $g(2012) = f(2012)-2011$, we get that $g(2012) = 1$
Therefore, our answer is $g(2012) = \boxed{1}$
Is my approachment fine? Are there any glitch or something wrong I got there?
