[proofplan] The forward implication is proved by induction on the number of points in the finite convex combination. The induction step separates the last coefficient: if the remaining total weight is positive, the first part is normalized to a smaller convex combination and then recombined with the last point using convexity; if the remaining total weight is zero, the whole combination is the last point. The reverse implication is obtained by applying the finite-combination property to two points with coefficients $t$ and $1-t$. [/proofplan]

[step:Prove convex sets contain all finite convex combinations]Assume that $C$ is convex. For each integer $m \geq 1$, let $P_m$ denote the assertion that for every choice of points $c_1,\dots,c_m \in C$ and scalars $\lambda_1,\dots,\lambda_m \in \mathbb{R}$ with $\lambda_i \geq 0$ for all $i$ and \begin{align*} \sum_{i=1}^m \lambda_i = 1, \end{align*} the vector \begin{align*} \sum_{i=1}^m \lambda_i c_i \end{align*} belongs to $C$. The assertion $P_1$ holds. Indeed, if $c_1 \in C$ and $\lambda_1 \geq 0$ with $\lambda_1 = 1$, then \begin{align*} \sum_{i=1}^1 \lambda_i c_i = c_1 \in C. \end{align*} Now fix an integer $m \geq 1$ and assume $P_m$. We prove $P_{m+1}$. Let $c_1,\dots,c_{m+1} \in C$ and let $\lambda_1,\dots,\lambda_{m+1} \in \mathbb{R}$ satisfy $\lambda_i \geq 0$ for every $i \in \{1,\dots,m+1\}$ and \begin{align*} \sum_{i=1}^{m+1} \lambda_i = 1. \end{align*} Define \begin{align*} \alpha := \sum_{i=1}^m \lambda_i. \end{align*} Then $\alpha \geq 0$ and $\alpha + \lambda_{m+1} = 1$. If $\alpha = 0$, then $\lambda_i = 0$ for every $i \in \{1,\dots,m\}$ because each $\lambda_i$ is nonnegative. Hence $\lambda_{m+1} = 1$, and therefore \begin{align*} \sum_{i=1}^{m+1} \lambda_i c_i = c_{m+1} \in C. \end{align*} Assume next that $\alpha > 0$. For each $i \in \{1,\dots,m\}$, define \begin{align*} \mu_i := \frac{\lambda_i}{\alpha}. \end{align*} Then $\mu_i \geq 0$ for every $i$, and \begin{align*} \sum_{i=1}^m \mu_i = \frac{1}{\alpha}\sum_{i=1}^m \lambda_i = 1. \end{align*} By the induction hypothesis $P_m$, the vector \begin{align*} y := \sum_{i=1}^m \mu_i c_i \end{align*} belongs to $C$. Since $\lambda_{m+1} = 1-\alpha$ and $\lambda_{m+1} \geq 0$, we have $\alpha \in [0,1]$. The convexity of $C$, applied to the two points $y,c_{m+1} \in C$ and the coefficient $\alpha \in [0,1]$, gives \begin{align*} \alpha y + (1-\alpha)c_{m+1} \in C. \end{align*} Substituting the definition of $y$ and using $\alpha \mu_i = \lambda_i$ for each $i \in \{1,\dots,m\}$, we obtain \begin{align*} \alpha y + (1-\alpha)c_{m+1} = \sum_{i=1}^{m+1} \lambda_i c_i. \end{align*} Thus the given $(m+1)$-point convex combination belongs to $C$, so $P_{m+1}$ holds. By induction, $P_m$ holds for every integer $m \geq 1$. Therefore every finite convex combination of points of $C$ belongs to $C$.[/step]

[guided]Assume that $C$ is convex. We want to prove that convexity under two-point combinations automatically gives convexity under any finite number of points. The natural way to do this is induction on the number of points. For each integer $m \geq 1$, let $P_m$ be the following assertion: whenever $c_1,\dots,c_m \in C$ and $\lambda_1,\dots,\lambda_m \in \mathbb{R}$ satisfy $\lambda_i \geq 0$ for all $i \in \{1,\dots,m\}$ and \begin{align*} \sum_{i=1}^m \lambda_i = 1, \end{align*} then \begin{align*} \sum_{i=1}^m \lambda_i c_i \in C. \end{align*} The base case $P_1$ says that a one-point convex combination of a point of $C$ lies in $C$. If $c_1 \in C$ and $\lambda_1 \geq 0$ with $\lambda_1 = 1$, then \begin{align*} \sum_{i=1}^1 \lambda_i c_i = c_1, \end{align*} and this belongs to $C$ by the choice of $c_1$. Now fix an integer $m \geq 1$ and assume $P_m$. We prove $P_{m+1}$. Choose points $c_1,\dots,c_{m+1} \in C$ and scalars $\lambda_1,\dots,\lambda_{m+1} \in \mathbb{R}$ such that every $\lambda_i$ is nonnegative and \begin{align*} \sum_{i=1}^{m+1} \lambda_i = 1. \end{align*} The key question is how to reduce an $(m+1)$-point combination to an $m$-point combination. We separate the last coefficient and define the total weight of the first $m$ terms by \begin{align*} \alpha := \sum_{i=1}^m \lambda_i. \end{align*} Because each $\lambda_i$ is nonnegative, $\alpha \geq 0$. Also, from the full sum condition, \begin{align*} \alpha + \lambda_{m+1} = 1. \end{align*} There are two cases. First suppose $\alpha = 0$. Since $\lambda_1,\dots,\lambda_m$ are nonnegative [real numbers](/page/Real%20Numbers) and their sum is zero, each of them must equal zero. Then the identity $\alpha + \lambda_{m+1} = 1$ gives $\lambda_{m+1} = 1$. Therefore the whole convex combination reduces to the last point: \begin{align*} \sum_{i=1}^{m+1} \lambda_i c_i = c_{m+1}. \end{align*} Since $c_{m+1} \in C$, the desired membership holds in this case. Now suppose $\alpha > 0$. We normalize the first $m$ coefficients so that they themselves form a convex combination. For each $i \in \{1,\dots,m\}$, define \begin{align*} \mu_i := \frac{\lambda_i}{\alpha}. \end{align*} Each $\mu_i$ is nonnegative because $\lambda_i \geq 0$ and $\alpha > 0$. Their sum is \begin{align*} \sum_{i=1}^m \mu_i = \frac{1}{\alpha}\sum_{i=1}^m \lambda_i = 1. \end{align*} Thus the induction hypothesis $P_m$ applies to the points $c_1,\dots,c_m$ and the coefficients $\mu_1,\dots,\mu_m$. It gives \begin{align*} y := \sum_{i=1}^m \mu_i c_i \in C. \end{align*} We now combine this new point $y \in C$ with the last point $c_{m+1} \in C$. Since $\lambda_{m+1} = 1-\alpha$ and $\lambda_{m+1} \geq 0$, the coefficient $\alpha$ lies in $[0,1]$. The defining convexity property of $C$ therefore gives \begin{align*} \alpha y + (1-\alpha)c_{m+1} \in C. \end{align*} Finally we compute this two-point convex combination: \begin{align*} \alpha y + (1-\alpha)c_{m+1} = \alpha\sum_{i=1}^m \mu_i c_i + \lambda_{m+1}c_{m+1}. \end{align*} Using $\alpha\mu_i = \lambda_i$ for each $i \in \{1,\dots,m\}$, this becomes \begin{align*} \alpha y + (1-\alpha)c_{m+1} = \sum_{i=1}^{m+1} \lambda_i c_i. \end{align*} Hence the original $(m+1)$-point convex combination belongs to $C$. This proves $P_{m+1}$. The base case and induction step prove, by mathematical induction, that $P_m$ holds for every integer $m \geq 1$. Thus every finite convex combination of points of $C$ belongs to $C$.[/guided]

[step:Recover convexity from two-point finite convex combinations] Assume conversely that every finite convex combination of points of $C$ belongs to $C$. To prove that $C$ is convex, let $x,y \in C$ and let $t \in [0,1]$. Define $m := 2$, $c_1 := x$, $c_2 := y$, $\lambda_1 := t$, and $\lambda_2 := 1-t$. Then $\lambda_1,\lambda_2 \geq 0$ and \begin{align*} \lambda_1+\lambda_2 = 1. \end{align*} By the assumed finite-combination property, \begin{align*} tx + (1-t)y = \lambda_1 c_1+\lambda_2 c_2 \in C. \end{align*} Since this holds for all $x,y \in C$ and all $t \in [0,1]$, the set $C$ is convex. Combining this with the forward implication proves the equivalence. [/step]