# Sets $E$ in $\mathbb{Z}$ such that any $l^2$ function with support on $E$ comes from Fourier of a continuous function

Are there infinite sets $E\subset\mathbb{Z}$ such that any $f\in l^2(\mathbb{Z})$ with support on $E$ comes from the Fourier transform of a continuous function on $\mathbb{T}$ ? If yes, is there a characterization of these sets?

If we change the group from $\mathbb{Z}$ to the free group with n generators $\mathbb{F_n}$, $n\geq 2$,then an infinite set $E$ with the property that $x_{i_1}x_{i_2}^{-1}x_{i_3}x_{i_4}^{-1}\ldots x_{i_{2k-1}}x_{i_{2k}}^{-1}\neq 1,\ \forall k\in\mathbb{N},\ \forall x_{i_1}\neq x_{i_2},\ x_{i_2}\neq x_{i_3},\ldots x_{i_{2k-1}}\neq x_{2k}$ (studied by Leinert, Bozejko) will satisfy the following:

Any $f\in l^2(\mathbb{F_n})$ with support on $E$ will be in the reduced $C^*$ algebra $C^*_r(\mathbb{F_n})$, and moreover $$||f||_\infty\leq 2||f||_2$$

[Edit:] By $||\cdot||_\infty$ I mean the operator norm of the corresponding convolutor.

-
One characterization of an amenable group $\Gamma$ is that $\| \lambda(f) \|_\infty = \| f \|_1$ for any non-negative valued function $f \in \ell^1\Gamma$. So no amenable groups will have such an infinite subset. Which then begs the question: What about non-amenable groups which do not contain a non-cyclic free group? –  Jesse Peterson Nov 13 '13 at 1:32

1) If you mean "Any $L^2$ function with Fourier coefficients support on $E$ must be continuous," then a set $E$ will have this property if and only if it is finite.
The fact that finite sets have this property is obvious. Conversely, assume that $E$ is infinite and let $\{a_n\}_{n=1}^{\infty}$ be an enumeration of your set. The function $f(x) = \sum_{n=1}^{\infty} \frac{1}{n} e(a_n x)$ will be in $L^2(\mathbb{T})$ but will have a singularity at $0$ and thus won't be continuous on $\mathbb{T}$. [This isn't quite precise since the function could conceivably agree almost everywhere with a continuous function, but more elaborate constructions along this line will work].
2) If you mean "Given any $\ell^2$ sequence $g(n)$ (for $n \in E$), there exists a continuous function $f$ on $\mathbb{T}$ such that $\hat{f}(n)=g(n)$ for $n \in E$, but may be arbitrary for $n \notin E$," then this is equivalent to $E$ being a $\Lambda(2)$ set.