Claim 1: Let $n\geq 1$. Then \[|u_{n+3}|\leq \max \{|u_{n+1}|, |u_n|\}\] Proof. Assume $u_{n+3}> 0$ (otherwise replace all $u_n$ with $-u_n$ and $f(x)$ with $-f(-x)$). Then $u_{n+2} > u_{n+1}$. Thus $u_{n+3}=f(u_{n+2})-f(u_{n+1})<u_{n+2}-u_{n+1}$. If $u_{n+2}\leq 0$ we get $u_{n+3}\leq -u_{n+1}\leq |u_{n+1}|$ which is good. Otherwise $u_{n+2}>0$, so $u_{n+1}>u_n$. Finally, \[u_{n+3}< u_{n+2}-u_{n+1}=f(u_{n+1})-f(u_n)-u_{n+1}<u_{n+1}-u_n-u_{n+1}=-u_n\leq |u_n|\]which finishes the proof. $\blacksquare$ This implies that the function $m(n)=\max \{|u_n|, |u_{n+1}|, |u_{n+2}|\}$ is non-increasing, so it approaches some limit $L$. Assume $L>0$ as otherwise we are done. We can assume that $m(n)\in [L, L+1]$ for all $n$ by changing indices. Lemma: There exists $\epsilon>0$ so that if $|x|, |y|\leq L+1$ and $x-y\geq L/2$ then \[f(x)-f(y)\leq x-y-\epsilon\] Proof. We can assume $x-y=L/2$ as $f(x)-x$ is decreasing. The function $L/2+f(y)-f(y+L/2)$ is positive and continuous (as $f$ is continuous, in fact, Lipschitz) on the compact interval $[-L+1, L/2+1]$ so it achieves a positive minimum value. $\blacksquare$ To finish, we now prove a strengthening of claim 1: Claim 2: Fix $\epsilon>0$ from the lemma. Then \[|u_{n+3}|\leq \max \{|u_{n+1}|-\epsilon, |u_n|-\epsilon, L/2\}\]for all $n\geq 1$. Proof. Assume $|u_{n+3}|\geq L/2$. As before, we can actually assume $u_{n+3}\geq L/2$. Thus $u_{n+2}>u_{n+1}$, and $u_{n+3}<u_{n+2}-u_{n+1}\implies u_{n+2}-u_{n+1}\geq L/2$. By the lemma, \[u_{n+3}\leq u_{n+2}-u_{n+1}-\epsilon\]If $u_{n+2}\leq 0$ this is at most $|u_{n+1}|-\epsilon$ as needed. Otherwise, we know $u_{n+1}>u_n$, and hence \[u_{n+3}< u_{n+1}-u_n-u_{n+1}-\epsilon = -u_n-\epsilon\leq |u_n|-\epsilon\]which finishes the proof. $\blacksquare$ As eventually $|u_n|<L+\epsilon$, claim 2 implies $m(n)<L$ eventually, a contradiction.