I am trying to prove that the set is convex: $$\{(x \in \mathbb{R}^n ) | \lambda_{\text{max}}(A(x))\le1, \boldsymbol{Tr}(B^TA(x))\ge 1\}$$ where $A(x)=A_0+x_1A_1+...+x_nA_n$ and $A_i \in \mathbb{S}^n$, $B \in \mathbb{R}^{n \times n}$
It's not very obvious to me which operation that preserves convexity to use for this proof. Any hints would be very helpful.
Note that $\text{tr}(B^TA(x)) = \sum_{j=1} ^n x_j \cdot \text{tr}(B^TA_j) + \text{Tr}(B^TA_0)$ which is affine in $x$. Hence, the subset defined by the rightmost constraint only is certainly convex (it's just a shifted half-plane).
For the leftmost constraint, you can use that $\lambda_{\text{max}}(A+B) \leq \lambda_{\text{max}}(A) + \lambda_{\text{max}}(B)$ (this is kind of non-trivial, look up Weyl's inequality) and $\lambda_{\text{max}}(tA) = t\lambda_{\text{max}}(A)$. Then indeed, $$ \lambda_{\text{max}} \bigg(A(tx_1 + (1-t)x_2) \bigg) = \lambda_{\text{max}} \bigg(tA(x_1) + (1-t)A(x_2) \bigg) \leq t\lambda_{\text{max}}A(x_1) + (1-t)\lambda_{\text{max}}A(x_2) \leq 1.$$
Your set is the intersection of the two, and hence also convex.
