Boolean-cube interpolation

Each factor in ${\lambda }_b$ is $X_i$ when $b_i=1$ and $1-X_i$ when $b_i=0$. Hence ${\lambda }_b$ is multilinear. At a Boolean vector $c$, every factor equals one if $c=b$; if $c\mathrm{\ne }b$, a differing coordinate supplies a zero factor. Thus ${\lambda }_b(c)$ is one for $b=c$ and zero otherwise.

For uniqueness consider a multilinear polynomial $h$ vanishing at all Boolean vertices. In dimension zero it is a constant with value zero, hence is zero. For positive dimension assume the assertion in dimension $n-1$ and write $h=A+X_{n}B$ with $A,B$ multilinear in the other variables. Its restrictions $h_0=A$ and $h_1=A+B$ vanish on that smaller cube, so both are zero by the induction hypothesis.

The displayed finite sum is multilinear and takes value $f(c)$ at $c$. When $n=0$ it is the single constant $f(())$. In particular the zero table extends to zero and the constant-one table extends to one.

The formal identity $h=(1-X_n)h_0+X_n{h}_1$ gives $h=0$. Induction proves the vanishing assertion in every dimension; applying it to the difference of two extensions proves uniqueness. All identities used only field addition and multiplication, with $0\mathrm{\ne }1$, so characteristic two is included.