Can the 0 element of the Fourier Algebra be represented as a coefficient function of two non-zero vectors?

Question

Consider an infinite discrete group $G$, and its associated Hilbert space $l^{2}(G)$.

For $t\in G$, let $\lambda(t):l^{2}(G)\to l^{2}(G)$ denote the map $[\lambda(t)x](s) = x(t^{-1}s)$. That is, $\lambda:G\to B(l^{2}(G))$ is the left regular representation of $G$.

Choose two functions $x,y\in l^{2}(G)$, and suppose that the function $f(t) = \langle \lambda(t)x,y\rangle$, where $\langle \cdot ,\cdot \rangle$ is the usual inner product on $l^{2}(G)$.

It seems not wildly impossible that if $f(t) = 0$ for every $t\in G$, then one of $x$ or $y$ is identically $0$.

But I can't prove it or find a counter-example. Before I spend any more time attempting to prove it, I thought I might check here.

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The concise version of the question is whether the $0$ element of the Fourier Algebra $A(G)$ can be represented as a coefficient function of two non-zero vectors.

Note that the norm of a function $f\in A(G)$ is defined by

$$\|f\|_{A(G)} = \text{inf}\|x\|\cdot\|y\|,$$ where the infimum is taken over all possible $x,y\in l^{2}(G)$ such that $f(t) = \langle\lambda(t)x,y\rangle$.

Answer 1

If $0\ne x\in l^2(G)$ is not cyclic for $\lambda$, i.e., the closed linear span of vectors $\lambda(t)x$ is a proper subspace in $l^2(G)$, then one can find $0\ne y\in l^2(G)$ for which $\langle \lambda(t)x,y\rangle=0$ $(\forall t\in G)$ although $x\ne 0\ne y$.

For a counterexample one can look in two dimensional case. Let $G=({\mathbb Z}_2, +)$. Then $l^2(G)={\mathbb C}^2$ and $$ \lambda(0)=\left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right),\qquad \lambda(1)=\left(\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right). $$ It is obvious that $x=\left(\begin{array}{c} 1 \\ 1\end{array}\right)$ is an eigenvector for $\lambda(t)$ $(t\in G)$, which means that $\lambda(0)x$ and $\lambda(1)x$ span onedimensional subspace. Let $y=\left(\begin{array}{c} 1 \\ -1\end{array}\right)$. Then $\langle \lambda(t)x,y\rangle=0$ for any $t\in G$.

EDIT. Ups! I have overlooked that $G$ has to be infinite. Hence the counteexample is not relevant. But the first part of the answer is correct, I guess.

EDIT 2. If $G$ is commutative, say $G=({\mathbb Z}, +)$, then $\{ \lambda(t); t\in G\}$ generates a commutative subalgebra of operators in $B(l^2(G))$. Such an algebra cannot be transitive, which means that it has a proper non-trivial invariant subspace. Hence, there exists a non-zero vector which is not cyclic for $\{ \lambda(t); t\in G\}$.