Please note that this is NOT a duplicate of Proving: The sum of piecewise linear convex functions is piecewise linear convex.
Definition. Let $\mathbf{c}_1, \dots, \mathbf{c}_m \in \mathbb{R}^n$ be column vectors, $d_1, \dots, d_m \in \mathbb{R}$, and let $f: \mathbb{R}^n \to \mathbb{R}$ be defined by $$f(\mathbf{x}) = \max_{i\in\{1, \dots, m\}}(\mathbf{c}^{\prime}_i\mathbf{x}+d_i)$$ where $\prime$ denotes the matrix transpose. Then $f$ is convex, and we call $f$ a piecewise linear convex function.
This is what I would like to prove:
Suppose $f_1, \dots, f_m: \mathbb{R}^n \to \mathbb{R}$ are piecewise linear convex and let $f = \sum_{i=1}^{m}f_i$. Then $f$ is piecewise linear convex.
My attempt:
Let $\mathbf{c}_{11}, \mathbf{c}_{21}, \dots, \mathbf{c}_{N_11}, \mathbf{c}_{12}, \dots, \mathbf{c}_{N_22}, \dots, \mathbf{c}_{N_mm} \in \mathbb{R}^n$ and $d_{11}, \dots , d_{N_mm} \in \mathbb{R}$. By definition of piecewise linear convex, we may write $$f_i(\mathbf{x}) = \max_{j\in \{1, \dots, N_m\}}(\mathbf{c}^{\prime}_{ji}\mathbf{x}+d_{ji})\text{, } \quad i = 1, \dots, m\text{.}$$ Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ and $\lambda \in [0,1 ]$. Then $$f(\mathbf{x})=\sum_{i=1}^{m}f_i(\mathbf{x})=\sum_{i=1}^{m}\max_{j\in \{1, \dots, N_m\}}\left[\mathbf{c}^{\prime}_{ji}\mathbf{x}+d_{ji} \right]\text{.}\tag{*}$$ I understand that I need to show that (*) is equal to $$\max_{k \in \{1, \dots, N\}}(\mathbf{c}_k^{\prime}\mathbf{x}+d_k)$$ for some $N$, $\mathbf{c}_1, \dots, \mathbf{c}_k \in \mathbb{R}^n$, and $d_1, \dots, d_k \in \mathbb{R}$, but I can't figure out how to do this. I don't understand how the answer given at Proving: The sum of piecewise linear convex functions is piecewise linear convex applies to this problem.
I have already read the official solution from the solutions manual, but do not understand it. This is exercise 1.2, from Introduction to Linear Optimization by Bertsimas and Tsitsiklis.