A convex 4-polytope has at most m(m−3)/2 facets

Fix a vertex $v$ of $Q$ and let $C$ be the convex hull of its other vertices. Since $v$ is extreme, $v\mathrm{\notin }C$. Writing $C$ as the continuous image of the compact coefficient simplex shows that $C$ is compact, so choose $q\in C$ minimizing $\mathrm{\parallel }v-q\mathrm{\parallel }$. For every $y\in C$ and $0<t\le 1$, $q+t(y-q)\in C$, hence $\mathrm{\parallel }v-q-t(y-q){\mathrm{\parallel }}^2\ge \mathrm{\parallel }v-q{\mathrm{\parallel }}^2$. Expanding, dividing by $t$, and letting $t\downarrow 0$ gives $(v-q)\cdot (y-q)\le 0$. Therefore $\mathrm{\ell }(x)=(v-q)\cdot x$ satisfies $\mathrm{\ell }(v)-\mathrm{\ell }(y)\ge \mathrm{\parallel }v-q{\mathrm{\parallel }}^2>0$ for every $y\in C$. Choose $\delta >0$ smaller than every gap $g_u=\mathrm{\ell }(v)-\mathrm{\ell }(u)>0$ over the other vertices. The section $Q_v=Q\cap \{\mathrm{\ell }=\mathrm{\ell }(v)-\delta \}$ is a cross-section of the cone generated by the directions $u-v$. Indeed, the segment $[v,u]$ meets it at $x_u=v+\frac{\delta }{g_u}(u-v)$, and if $x={\lambda }_{v}v+{\sum }_{u\mathrm{\ne }v}{\lambda }_{u}u\in Q_v$, then ${\mu }_u={\lambda }_u{g}_u\mathrm{/}\delta$ are nonnegative, sum to $1$, and give $x={\sum }_{u\mathrm{\ne }v}{\mu }_u{x}_u$. Thus $Q_v=\mathrm{conv}\{x_u:u\mathrm{\ne }v\}$. The cone is four-dimensional because the vertex differences span ${\mathbb{R}}^4$, so its transverse section is three-dimensional. A face of this cone is generated by the directions of the corresponding face of $Q$ through $v$: a supporting linear inequality of the cone restricts to a supporting inequality of $Q$ tight at $v$, and conversely the facets active at $v$ define the cone. Therefore its extreme rays correspond to edges at $v$, and its three-dimensional faces correspond to facets of $Q$ containing $v$. A facet of $Q$ not containing $v$ misses the section because all its vertices lie below the section level. Intersecting cone faces with the transverse section lowers dimension by one, so the vertices of $Q_v$ correspond to edges of $Q$ at $v$ and its facets correspond to facets of $Q$ containing $v$. Write $d_v,t_v$ for the numbers of vertices and facets of $Q_v$.

Choose a facet $F_0$ of $Q_v$. A viewpoint just outside a relative-interior point of $F_0$ gives a Schlegel projection onto its plane: each ray through an interior point of $F_0$ enters $Q_v$ there and, by convexity and boundedness, has a unique exit point on the opposite boundary. The set of such viewpoints contains an open region. The exit points cover $F_0$; uniqueness makes projected facet interiors disjoint, and shared edges project to shared boundaries. Choose the viewpoint generically within this region so projected vertices are distinct and no three projected vertices are collinear. A pair coincidence forces the viewpoint onto the line through that pair; collinearity of three projected vertices forces it into the plane through those three noncollinear vertices. These finitely many proper algebraic sets cannot exhaust an open region: each lies in the zero set of a nonzero polynomial, and a nonzero polynomial cannot vanish on an open set (in one variable it has finitely many roots; induction on the variables applied to its coefficient polynomials proves the claim). Triangulate each projected convex polygonal facet by joining one vertex to its nonadjacent vertices. This yields a triangulation of $F_0$ with $d_v$ sites and $k_v$ hull sites, where $k_v$ is the number of vertices of $F_0$ and $k_v\ge 3$.

There are $t_v-1$ bounded facets in the Schlegel triangulation, and each contributes at least one triangle, so $t_v-1\le T_v$. By [F1], $T_v=2d_v-2-k_v$. Therefore $t_v\le 2d_v-1-k_v\le 2d_v-4$.

Count vertex-facet incidences. Every facet of a four-dimensional polytope has at least four vertices, so $4f_3\le {\sum }_v{t}_v$. By step 3.1 and the fact that each edge has two endpoints, ${\sum }_v{t}_v\le 2{\sum }_v{d}_v-4m=4f_1-4m$.

The graph of $Q$ is simple, so $f_1\le \left(\genfrac{}{}{0px}{}{m}{2}\right)$. Step 4.1 yields $4f_3\le 4f_1-4m$, hence $f_3\le f_1-m\le \left(\genfrac{}{}{0px}{}{m}{2}\right)-m=m(m-3)\mathrm{/}2$.