Let $X$ be a Hausdorff locally convex topological vector space. Suppose $X_0 \subset X$ is nonempty convex set, $g:\; X\to \mathbb R^m$ is a convex vector function (each component $g_i(x): X\to \mathbb R$ is convex function, $i=1,2,\ldots,m$). Prove that, if there is no $x\in X_0$ such that $g(x)<0$ then there exists $\lambda\in \mathbb R_{+}^m\setminus\{0\}$ such that $\langle \lambda, g(x)\rangle \geq 0$ for all $x\in X_0$.
Here is my attempt: Suppose that there is no $x\in X_0$ such that $g(x)<0$. For each $x_0\in X_0$, define
$$V(x)=\{\beta \in \mathbb R^m:\; g(x)<\beta\}\; \mbox{ and }\; V=\cup_{x\in X_0}V(x).$$ where $g(x)<\beta$ means $g_i(x)<\beta_i$ for each $i=1, \ldots, m$. Assume that $\beta^{(1)}\in V(x_1)$, $\beta^{(2)}\in V(x_2)$ then for $\lambda \in (0,1)$ we have $$g(\lambda x_1 + (1-\lambda)x_2)\leq \lambda g(x_1)+(1-\lambda)g(x_2)\leq \lambda \beta^{(1)}+(1-\lambda)\beta^{(2)}.$$ Thus $\lambda \beta^{(1)}+(1-\lambda)\beta^{(2)} \in V(\lambda x_1 + (1-\lambda)x_2)$ and since $\lambda x_1 + (1-\lambda)x_2 \in X_0$, we have $V$ is convex. $V\neq \emptyset$ because of the nonemptyness of $X_0$ [Take $x_0\in X_0$, and $\beta^{(0)}=g(x_0)+(1, \ldots, 1) \in V$]. By the hypothesis, $0\notin V$. Applying Separation Theorem, there exists $\lambda\in \mathbb R^m\setminus\{0\}$ such that
$$\langle \lambda, \beta\rangle\leq 0, \mbox{ for all } \beta\in V.$$ Noting that if $\beta\in V$ then $\beta +(p,\ldots,p) \in V$ for any $p>0$, from above inequality we can imply $\lambda \geq 0$, i.e., $\lambda_i \geq 0$ for all $i=1, \ldots, m$. On the other hand for each $x\in X_0$, let $\beta=g(x)+(1/n,1/n, \ldots, 1/n) \in V$, we have $$\langle \lambda, g(x)\rangle + \sum_{i=1}^m \frac{\lambda_i}{n} \geq 0.$$ Let $n\to \infty$ we conclude that $\langle \lambda, g(x)\rangle \geq 0$, for each $x\in X_0$. QED.
Is my proof right? Do we need any assumption on interior in above proof when using Separation Theorem?
Thanks in advance.
The proof looks good. You do not need any assumption on the interior as you are separating a point from a convex set. You would need a non-emptyness condition on the interior if you would like to separate two convex sets in infinite-dimensional spaces.