# Can we expect $\left\||f|^{2}f-|g|^{2}g\right\|\leq C ||f-g||$ in the Banach algebra $A(\mathbb T)$ ?

Let $f\in L^{1}(\mathbb T)$ and define the Fourier coefficient of $f$ : $\hat{f}(n)=\frac{1}{2\pi} \int _{-\pi}^{\pi} f(t) e^{-int} dt; (n\in \mathbb Z)$.Consider the space, $$A(\mathbb T):= \{f\in L^{1}(\mathbb T): \hat{f}\in \ell^{1}(\mathbb Z), \ \text {that is,} \ \sum_{n\in \mathbb Z} |\hat{f}(n)| < \infty \}.$$ $A(\mathbb T)$ is normed by the $L^{1}-$ norm on $\mathbb Z$: $$||f||= \sum_{n\in \mathbb Z} |\hat{f}(n)|; \ \text {for} \ f\in A(\mathbb T).$$ We also note that $A(\mathbb T)$ is a Banach algebra under pointwise addition and multiplication. Let $f_{0}\neq 0 \in A(\mathbb T)$ and fix it; and take $M= 2||f_{0}||$ and put $B_{M}= \{f\in A(\mathbb T): ||f||\leq M \}.$

Let $f, g \in B_{M}.$

My Question: Can we expect $\left\||f|^{2}f-|g|^{2}g\right\|\leq C ||f-g||$ , where $C$ is some constant ? If yes, what can we say about $C$ ?

I guess, the trivial relation: $|f|^{2}f-|g|^{2}g=(f-g)|f|^{2}+g(|f|^{2}-|g|^{2})$; may be useful.

Thanks,

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The Fourier transform is a *-homomorphism for the natural adjoint on $\ell^1(\mathbb Z)$ and so if $f$ is the transform of $a\in\ell^1(\mathbb Z)$ then $|f|^2f$ is the transform of $a^*aa$. Similarly let $g$ and $b$ be related. As $\|a\|,\|b\| \leq M$ we perform a simply triangle-inequality argument, which is valid in any Banach $*$-algebra: \begin{align*} \| a^*aa - b^*bb\| &= \| (a^*-b^*)aa + b^*(a-b)a + b^*b(a-b) \| \\ &\leq \|a-b\|\|aa\| + \|b\| \|a\| \|a-b\| + \|b^*b\| \|a-b\| \leq 3M^2\|a-b\|. \end{align*}
Your constant depends on $a$ and $b$. The point is a uniform constant. But maybe your estimates can be made sharp disproving the conjecture of the OP. – Marc Palm Mar 6 '14 at 14:12
But that's not what the question asked: $f_0$ is fixed, this gives $M$ and then the OP specifically asks only for $f,g\in B_M$. If it was meant to be uniform, then why both with the $B_M$ condition? – Matthew Daws Mar 6 '14 at 14:28
Although, I must say that the whole $M=2\|f_0\|$ thing is very mysterious... – Matthew Daws Mar 6 '14 at 14:30
Ah okay, I didn't carefully read, where $f$ and $g$ are living! – Marc Palm Mar 6 '14 at 15:18
@MD; thanks; my motivation was to prove contraction in the space $C([0, T];M^{p,1})$; for detail, please have look on my previous question ; on the torus modulation space coincide with $A(T)$; thanks for the attention; – Inquisitive Mar 6 '14 at 15:44