Let $(f_i)_{i \in \{1,2\}} : X \to \mathbb R$ denote a pair of objective functions and $X \subseteq \mathbb R^2$ the set of admissible controls. Let \begin{align} X^e = \{x \in X \mid \nexists y \in X: f_i(y) \geq f_i(x) \forall i, f_i(y) > f_i(x) \exists i\} \end{align} denote the set of efficient controls. Further let $P = \{(f_1(x),f_2(x)) \mid x \in X^e\}$ denote the Pareto frontier.
Now consider the weighted problem. For $\lambda \in [0,1]$ let $f_\lambda(x) = \lambda f_1(x) + (1-\lambda)f_2(x)$ and suppose that $f_\lambda(x)$ is concave. Let $x^*(\lambda)$ solve \begin{align} x^*(\lambda) = \arg\max_{x \in X}f_\lambda(x). \end{align} I was wondering under what conditions the Pareto frontier coincides with allocations given by the maximizers of the weighted problem, i.e., \begin{align} P \stackrel{?}{=} \{(f_1(x^*(\lambda)), f_2(x^*(\lambda))) \mid \lambda \in [0,1]\}. \end{align}
I will assume for simplicity that you want to minimize(and not maximize) the function $f$. I will also assume that $f_1,f_2$ are convex (in the maximization case, they would be concave), and that the set $X$ is convex. Then, it can be shown that
$$\{(f_1(x^*(\lambda)), f_2(x^*(\lambda))) \mid \lambda \in [0,1]\}= \{x\in X\mid \nexists y \in X: f_i(y)<f_i(x) \forall i \},$$ that is, the weakly Pareto frontier. Thus, in order to have the equality you want, you need to have that Pareto = weak Pareto frontier.