Question : In $\mathbb{R}^n$, consider an inner product $(\ ,\ )$. Here any linear map $L$ has the form $L(x)=(V,\ x)$ for some $V$
When $\|\ \|,\ \|\ \|^\ast$ are norms in dual relation, then we have their unit circles $C,\ C^\ast$ (For any $x\in C$, there is $x^\ast \in C^\ast$ s.t. $(x^\ast, x)=1\geq (x^\ast,y)$ for all $y\in C$).
Here the linear map $(x^\ast,\ )$ is called a support map at $x\in C$.
Further we have the set of quadratic forms $$A_C = \{ nL^2| L =(x^\ast,\ ),\ x^\ast \in C^\ast \}$$
Prove that any element $Q$ in the convex hull $ \overline{A}_C$ of the set $A_C$ has the convex combination of at most $\frac{n(n+1)}{2} +1$ elements in $A_C$ i.e. $$ Q = \sum_{i=1}^N\ a_i nL_i^2,\ a_i>0,\ \sum_i\ a_i=1,\ N\leq \frac{n(n+1)}{2} +1,\ nL_i^2\in A_C$$
Proof : Element in convex hull has $Q = \sum_{i=1}^N\ a_i nL_i^2$ in general where $\sum_i\ a_i=1,\ a_i>0$. We have a claim that we must find the upper bound for $N$.
Since $Q$ is a quadratic form, then there is a positive definite symmetric matrix $S$ s.t. $Q(x)= (Sx,x)$.
Hence there is a diagonalization $S = {\rm diag}\ (\lambda_1^2,\cdots,\lambda_n^2)$ wrt orthonormal set $\{e_i\}$.
When $L_i=(t_ie_i,\ )$ for some $t_i>0$ s.t. $nL_i^2\in A_C$, then $$ Q(x) = \sum_{i=1}^n\ \frac{\lambda_i^2}{nt_i^2} nL_i^2 $$
This may be a wrong direction since there is at most $n$ coefficients. How can we prove this ?
[Add] The set of all quadratic forms in $\mathbb{R}^n$ forms a vector space of dimension $\frac{n(n+1)}{2}$. Remaining is followed from the theorem stated by user804886.