For an engineering problem, I need to be able to bound from above the maximum eigenvalue of the following quantity: $$\lambda_\max\left(\sum^{N}_{i=1} w_i A_i\right)$$ where $w_i \in \mathbb{R}, i= 1,\ldots,N$ are some real (not positive in general) coefficients, $A_i \in \mathbb{R}^{n\times n}, i = 1, \ldots, n$ are symmetric, indefinite in general (actually, their leading eigenvalue can be positive or negative; the others are all negative), full-rank matrices, for which I already know (as given prior information) that: $$\lambda_\max\left(\sum^{N}_{i=1} A_i\right) < C, C > 0$$ I was wondering if there is any known way to make any norm of the sequence $\{w_i\}^N_{i=1}$ appear in an upper bound, i.e., to show that $$ \lambda_\max\left(\sum^{N}_{i=1} w_i A_i\right) \lesssim f(\|w\|_p,C)$$ with $\|w\|_p$ the $p$-norm of the vector $w = \{w_i\}^N_{i=1}$ and $C$ given above.
Of course for $A_i$ symmetric and positive-semidefinite matrices, and positive $w$ the solution should be easy (by saturating the sum with $\|w\|_{\infty}$).
No. Consider the matrix $$ A_1 = \begin{bmatrix} 1 & 0 \\ 0 & u \end{bmatrix} $$ where $u < 0$, and the one-term sum $\omega_1 A_1$. In this case, $\lambda_{max} = 1$, so we can pick $C = 1$.
Now picking $\omega_1 = -1$, we get $$ \lambda_{max}(\omega_1 A_1) = -u, $$ where $u$ is an arbitrary negative number. So you can certainly make it larger than $f(1, 1)$ no matter what $f$ you pick.