Truncated function condition and boundedness.

Question

Let $f:\mathbb{R}\to\mathbb{R}$ a function with $f(0)=0$, $f(x)\geq 0$ and $\lim_{|x|\to\infty} f(x)=\infty$. Define a new function on $\mathbb{R}\times Y$ by $g(x,y)=f(y-x)$ where $Y$ is a bounded and closed subset of $\mathbb{R}$. Define also the truncated value of $x$ at $M>0$ by $$x^{*}=\begin{cases} -M & x<-M \\ x & -M\leq x\leq M \\ M & x>M \end{cases}$$ There are no continuity assumptions over $f$. I want to prove that under this assumptions there exists a value $M>0$ such that for all $x,y\in \mathbb{R}\times Y$ $$g(x^{*}, y)\leq g(x,y)$$ I already proved the reciprocal of this result, but I'm stuck at this point. Any sugestion for the problem?