No translation-invariant normalised functional on all real functions on the integers

There is no $\mathbb{R}$-linear functional $m$ on the space of all real-valued functions on $\mathbb{Z}$ that is both translation-invariant and normalised.

Precisely, no $m$ satisfies both

• $m\bigl(n \mapsto f(n-1)\bigr) = m(f)$ for every $f : \mathbb{Z} \to \mathbb{R}$, and
• $m(\mathbf{1}) = 1$, where $\mathbf{1}$ is the constant function with value $1$.

Invariance is imposed only for the one-step shift. For a linear functional on $\mathbb{Z}$ that is equivalent to invariance under every translation, since the shift by one generates them all, so nothing is lost by stating the weaker hypothesis — and the theorem is correspondingly stronger.

The domain is the full function space, with no boundedness condition, and $m$ is required only to be linear: no positivity, continuity or norm condition appears. The statement is therefore stronger than the corresponding failure for an invariant mean, which would in addition be positive — not even a bare linear functional survives here.

The obstruction is the unbounded function $n \mapsto n$. Its translate $n \mapsto n-1$ differs from it by the constant function $\mathbf{1}$, so invariance and linearity together force $m(\mathbf{1})$ to be $0$, contradicting normalisation.

This is the reason an invariant mean is defined on $\ell^\infty(G)$ rather than on all functions: widening the domain does not merely cost positivity, it makes the notion unsatisfiable. The argument shown here needs a function whose translate differs from it by a nonzero constant, which $\mathbb{Z}$ supplies; it is not a claim about every infinite group.