# $f,g , |f|f, |g|g \in A(\mathbb R) \ \text{(Banach algebra)} \implies \left\|f|f|- g|g|\right\|\leq C \left \|f-g\right \|$?

Let $f\in L^{1}(\mathbb R)$ and it Fourier transform, $\hat{f} (y) : = \int _ {\mathbb R} f(x) e^{-2\pi i x\cdot y} dx ; y \in \mathbb R ;$ and consider Fourier algebra $$A(\mathbb R):= \{f\in L^{1}(\mathbb R): \hat{f}\in L^{1}(\mathbb R) \}$$ $A(\mathbb R)$ is normed by: $$\left\|f\right\|:= \left\|\hat{f}\right\|_{L^{1}(\mathbb R)}=\int_{\mathbb R}|\hat{f}(\xi)| d\xi; \ (f\in A(\mathbb R)).$$ We note that $A(\mathbb R)$ is a Banach algebra under point wise addition and multiplication.

Let $M>0$ and fix it; and consider $B_{M}= \{f\in A(\mathbb R): f|f|\in A(\mathbb R) \ \text{and} \ ||f||\leq M \}.$

My Question: Can we expect, $\left\|f|f|- g|g|\right\|\leq C \left \|f-g\right \|$; for $f,g \in B_{M}$ , where $C$ is some Constant ? If yes, what can we say about $C$ ? If not, can we produce counter example ?

Thanks,

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$f|f|\in L^1$ simply means $f\in L^2$ doesn't it? – username Oct 28 '14 at 6:55