Lemma 2

*Let H and $H'$ be two distinct hyperplanes of E, and f, $f'$ two affine functions on E such that H (resp. $H'$) consists of the points a in E such that $f(a) = 0$ (resp. $f'(a) = 0$). Finally, let L be a hyperplane of E. Assume that one of the following hypotheses is satisfied: a) The hyperplanes H, $H'$ and L are parallel. b) The hyperplanes H and $H'$ are not parallel, and $H \cap H' \subset L$. Then there exist real numbers $\lambda, \lambda'$, not both zero, such that L consists of those points $a \in E$ at which the affine function $g = \lambda . f + \lambda' . f'$ vanishes.

The lemma being trivial when $L = H$, we assume that there exists a point $a$ in L with $a \notin H$. Put $\lambda = f'(a), \lambda' = -f(a)$ and

then $\lambda' \neq 0$ since $a \notin H$; moreover, since $H \neq H'$, there exists $b \in H$ such that $b \notin H'$, so that $f(b) = 0, f'(b) \neq 0$, and thus $g(b) = -f(a).f'(b)$ is non-zero. The set $L_1$ of points where the affine function $g \neq 0$ vanishes is then a hyperplane of E; we have $g(a) = 0$, so $a \in L_1$.

a) Assume that H and $H'$ are parallel: g and f both vanish at every point of $L_1 \cap H$, hence so does $f'$ since $\lambda' \neq 0$; thus every point of $L_1 \cap H$ belongs to $H'$; but since H and $H'$ are parallel and distinct, they are disjoint, so $L_1 \cap H = \varnothing$, and Lemma 1 shows that $L_1$ is parallel to H. Since $a \in L$ and $a \in L_1$, we thus have $L = L_1$.

b) Assume that H and $H'$ are not parallel: by Lemma 1, there is a point c in $H \cap H'$; we give E the vector space structure obtained by taking c as the origin. Then $H \cap H'$ is a vector subspace of $E$ of codimension 2, and since $a \notin H$, the vector subspace $M$ of $E$ generated by $H \cap H'$ and $a$ is a hyperplane; since $H \cap H' \subset L \cap L_1$ and $a \in L \cap L_1$, we have $M \subset L \cap L_1$, hence $M = L = L_1$. Q.E.D.