Is contraction of $L^{2}$ function is again $L^{2}$ function?

Question

Suppose that $f\in L^{2}(\mathbb R).$ Let $g:\mathbb R \to \mathbb C$ be a contraction of $f$, that is, $|g(x)-g(y)|\leq |f(x)-f(y)|$ for all $x, y\in \mathbb R.$

My Question is: Is it true that $g\in L^{2}(\mathbb R)$ ?

Answer 1

Let $f(x)=1/x^2$ for $|x|>1$ and 1 for $|x|\leq1$. In the sense of your definition $g(x)\equiv1$ is a contraction of $f$, but is not in $L^2(\mathbb{R})$.