[proofplan] We verify convexity directly from the definition. Given two points $x,y \in C$ and a parameter $t \in [0,1]$, convexity of $C$ ensures that the convex combination $(1-t)x+ty$ lies in the domain of every $f_i$. Convexity of each $f_i$ gives one inequality; multiplying by the nonnegative coefficient $a_i$ preserves the inequality direction, and summing the finitely many resulting inequalities gives the desired convexity inequality for $f$. [/proofplan] [step:Apply convexity of each summand at an arbitrary convex combination] Let $x,y \in C$ and let $t \in [0,1]$ be arbitrary. Since $C$ is convex, the point \begin{align*} z=(1-t)x+ty \end{align*} belongs to $C$. Therefore $f_i(z)$ is defined for every $i \in \{1,\ldots,m\}$. For each $i \in \{1,\ldots,m\}$, the function $f_i:C\to\mathbb{R}$ is convex, so applying the definition of convexity to the points $x,y \in C$ and the parameter $t \in [0,1]$ gives \begin{align*} f_i((1-t)x+ty) \le (1-t)f_i(x)+t f_i(y). \end{align*} [guided] We begin with the exact data needed to test convexity of $f$. Let $x,y \in C$ and let $t \in [0,1]$ be arbitrary. To prove that $f$ is convex, we must prove the inequality \begin{align*} f((1-t)x+ty) \le (1-t)f(x)+t f(y). \end{align*} The expression on the left is meaningful only if $(1-t)x+ty$ lies in the domain of $f$, namely in $C$. This is exactly where the convexity of the set $C$ is used: since $x,y \in C$ and $t \in [0,1]$, the point \begin{align*} z=(1-t)x+ty \end{align*} belongs to $C$. Now each summand $f_i:C\to\mathbb{R}$ is convex by hypothesis. Applying the definition of convexity of $f_i$ to the same points $x,y \in C$ and the same parameter $t \in [0,1]$ yields, for every $i \in \{1,\ldots,m\}$, \begin{align*} f_i((1-t)x+ty) \le (1-t)f_i(x)+t f_i(y). \end{align*} This produces one valid real inequality for each index $i$. [/guided] [/step] [step:Multiply by nonnegative coefficients and sum the inequalities] Fix $i \in \{1,\ldots,m\}$. Since $a_i \ge 0$, multiplying the preceding inequality by $a_i$ preserves the inequality direction: \begin{align*} a_i f_i((1-t)x+ty) \le a_i\bigl((1-t)f_i(x)+t f_i(y)\bigr). \end{align*} Using distributivity in $\mathbb{R}$, this becomes \begin{align*} a_i f_i((1-t)x+ty) \le (1-t)a_i f_i(x)+t a_i f_i(y). \end{align*} Summing these inequalities over the finite index set $\{1,\ldots,m\}$ gives \begin{align*} \sum_{i=1}^m a_i f_i((1-t)x+ty) \le \sum_{i=1}^m \bigl((1-t)a_i f_i(x)+t a_i f_i(y)\bigr). \end{align*} By finite additivity and distributivity of sums in $\mathbb{R}$, \begin{align*} \sum_{i=1}^m \bigl((1-t)a_i f_i(x)+t a_i f_i(y)\bigr)=(1-t)\sum_{i=1}^m a_i f_i(x)+t\sum_{i=1}^m a_i f_i(y). \end{align*} By the definition of $f:C\to\mathbb{R}$, we have \begin{align*} \sum_{i=1}^m a_i f_i((1-t)x+ty)=f((1-t)x+ty). \end{align*} Also, \begin{align*} \sum_{i=1}^m a_i f_i(x)=f(x) \end{align*} and \begin{align*} \sum_{i=1}^m a_i f_i(y)=f(y). \end{align*} Substituting these three identities into the summed inequality gives \begin{align*} f((1-t)x+ty) \le (1-t)f(x)+t f(y). \end{align*} [/step] [step:Conclude convexity of the finite nonnegative linear combination] The points $x,y \in C$ and the parameter $t \in [0,1]$ were arbitrary. Hence the inequality \begin{align*} f((1-t)x+ty) \le (1-t)f(x)+t f(y) \end{align*} holds for all $x,y \in C$ and all $t \in [0,1]$. By the definition of convexity, $f$ is convex on $C$. [/step]