Let $\lambda_1, \lambda_2 >0$ be coefficients and $\mathbf{c},\; \mathbf{a_1}, \; \mathbf{a_2} \in \mathbb{R}^n$ be given vectors.
I am wondering whether
$$f(\theta) \; = \; \inf_{\mathbf{x}, \mathbf{y}}\{ \mathbf{x}^\top \mathbf{a_1} + \mathbf{y}^\top \mathbf{a_2} \ : \ \mathbf{x} + \mathbf{y} = \theta\cdot \mathbf{c}, \; \lVert \mathbf{x} \rVert \leq \lambda_1, \; \lVert \mathbf{y} \rVert \leq \lambda_2 \}$$
is an affine function on a given compact set $\Theta \subset \mathbb{R}$. The coefficients $\mathbf{c}$ satisfy $\lVert \theta\cdot \mathbf{c}\rVert \leq \lambda_1 + \lambda_2$ for all $\theta \in \Theta$ which implies $f$ never evaluates to $\infty$ on $\Theta$.
Why I think it might be affine
The function $f(\theta)$ can be written with variable substitution as (for simplicity use $\lambda_1 = \lambda_2= 1$): $$ f(\theta) = \theta\cdot \mathbf{c}^\top \mathbf{a_2} + \min_{\mathbf{x}}\{ \mathbf{x}^\top \mathbf{a} \ : \ \lVert \mathbf{x} \rVert \leq 1, \ \lVert \theta \cdot \mathbf{c} - \mathbf{x}\rVert \leq 1 \} $$ where $\mathbf{a} = \mathbf{a_1} - \mathbf{a_2}$. Since the first term is linear already, ignoring it (as we want to prove linearity) gives us $$ f(\theta) = \min_{\mathbf{x}}\{ \mathbf{x}^\top \mathbf{a} \ : \ \lVert \mathbf{x} \rVert \leq 1, \ \lVert \theta \cdot \mathbf{c} - \mathbf{x}\rVert \leq 1 \} $$ For fixed $\theta$, this problem is a linear optimization over the intersection of two unit balls. The first ball is centered at the origin while the second ball's center is $\theta \cdot \mathbf{c}$. So, varying $\theta$ linearly 'shifts' the second ball. But I cannot see whether if this implies that $f(\theta)$ is linear in $\theta$.