For any ${\displaystyle x\in \mathrm {Conv} (S)}$, represent ${\displaystyle x=\sum _{n=1}^{N}w_{n}q_{n}}$ for some ${\displaystyle q_{1},...,q_{N}\in S}$, then ${\displaystyle x\in \mathrm {Conv} (\{q_{1},...,q_{N}\})}$, and we use the lemma.

The second part reduces to the first part by "lifting up one dimension", a common trick used to reduce affine geometry to linear algebra, and reduce convex bodies to convex cones.

Explicitly, let ${\displaystyle S\subset \mathbb {R} ^{d}}$, then identify ${\displaystyle \mathbb {R} ^{d}}$ with the subset ${\displaystyle \{w\in \mathbb {R} ^{d+1}:w_{d+1}=1\}}$. This induces an embedding of ${\displaystyle S}$ into ${\displaystyle S\times \{1\}\subset \mathbb {R} ^{d+1}}$.

Any ${\displaystyle x\in S}$, by first part, has a "lifted" representation ${\displaystyle (x,1)=\sum _{n=1}^{N}w_{n}(q_{n},1)}$ where at most ${\displaystyle d+1}$ of ${\displaystyle w_{n}}$ are nonzero. That is, ${\displaystyle x=\sum _{n=1}^{N}w_{n}q_{n}}$, and ${\displaystyle 1=\sum _{n=1}^{N}w_{n}}$, which completes the proof.

This is trivial when ${\displaystyle N\leq d}$. If we can prove it for all ${\displaystyle N=d+1}$, then by induction we have proved it for all ${\displaystyle N\geq d+1}$. Thus it remains to prove it for ${\displaystyle N=d+1}$. This we prove by induction on ${\displaystyle d}$.

Base case: ${\displaystyle d=1,N=2}$, trivial.

Induction case. Represent ${\displaystyle x=\sum _{n=1}^{d+1}w_{n}q_{n}}$. If some ${\displaystyle w_{n}=0}$, then the proof is finished. So assume all ${\displaystyle w_{n}>0}$

If ${\displaystyle \{q_{1},...,q_{d}\}}$ is linearly dependent, then we can use induction on its linear span to eliminate one nonzero term in ${\displaystyle \sum _{n=1}^{d}{\frac {w_{n}}{w_{1}+\cdots +w_{d}}}q_{n}}$, and thus eliminate it in ${\displaystyle x=\sum _{n=1}^{d+1}w_{n}q_{n}}$ as well.

Else, there exists ${\displaystyle (u_{1},...,u_{d})\in \mathbb {R} ^{d}}$, such that ${\displaystyle \sum _{n=1}^{d}u_{n}q_{n}=q_{d+1}}$. Then we can interpolate a full interval of representations:

$${\displaystyle x=\sum _{n=1}^{d}(w_{n}+\theta w_{d+1}u_{n})q_{n}+(1-\theta )w_{d+1}q_{d+1};\quad \theta \in [0,1]}$$

If ${\displaystyle w_{n}+w_{d+1}u_{n}\geq 0}$ for all ${\displaystyle n=1,...,d}$, then set ${\displaystyle \theta =1}$. Otherwise, let ${\displaystyle \theta }$ be the smallest ${\displaystyle \theta }$ such that one of ${\displaystyle w_{n}+\theta w_{d+1}u_{n}=0}$. Then we obtain a convex representation of ${\displaystyle x}$ with ${\displaystyle \leq d}$ nonzero terms.