Let $X$ be a normed $\mathbb R$-vector space, $S\subseteq X\times\mathbb R$ and $$K=\left\{\sum_{i=1}^k\lambda_ix_i:k\in\mathbb N,x_i\in S,\lambda_i\ge0\right\}.$$
In a lecture note, where the case $X=\mathbb R^d$ was discussed, I've seen the claim that if $(0,1)$ is an interior point of $S$, then $$(0,1)=\sum_{i=1}^{d+1}\lambda_ix_i$$ for some linearly indepednent $x_1,\ldots,x_{d+1}\in S$ and $\lambda_1,\ldots,\lambda_{k+1}\ge 0$.
Question 1: How can we show this? And, most importantly, can we obtain a similar conclusion in the case of a general Banach or Hilbert space?
Question 2: Moreover, it is claimed that if $(0,1)$ is not an interior point of $S$, then there is a nonzero $w\in X\times\mathbb R$ such that $\langle w,z\rangle\le0$ for all $z\in S$ and $w_2\ge0$. We clearly want to assume that $X$ is a Hilbert space here, but even when $X=\mathbb R^d$, I don't get how this result follows.
I guess the case $X=\mathbb R^d$ is related to the Carathéodory theorem; though I don't know if the claim does actually from it. However, I'm mostly interested in the case of an infinite-dimensional Banach or Hilbert space.
EDIT: Maybe we can show that for an interior point of $K$, every $\lambda_i$ must be positive?
Q1: This holds in arbitrary Banach spaces. Let $y \in K$ be given. Thus, $y$ can be written in the form $$ y = \sum_{i=1}^k \lambda_i x_i$$ with $\lambda_i > 0$ and $x_i \in S$. W.l.o.g. we assume that $k$ is as small as possible. If the vectors $x_1,\ldots,x_k$ would not be linearly independent, we have $$ 0 = \sum_{i = 1}^k \alpha_i x_i$$ for some nonzero $\alpha \in \mathbb R^k$. But then we can choose an appropriate $t \in \mathbb R$ such that $$ y = \sum_{k=1}^k (\lambda_i + t \alpha_i) x_i, $$ $\lambda_i + t \alpha_i \ge 0$ for all $i$ and $\lambda_j + t \alpha_j = 0$ for at least one $j$. Thus, we can remove all vanishing terms from the sum and we get a representation of $y$ with fewer than $k$ vectors, which is not possible. Hence, $x_1,\ldots,x_k$ are linearly independent.
[Note that $y$ does not need to be an interior point.]
Q2: This fails in infinite dimensional inner product spaces. Indeed, if $S$ is a dense, proper subspace of $X \times \mathbb R$, then $K = S$, the interior of $S$ is empty and $\langle w,z\rangle \le 0$ for all $z \in S$ implies $w = 0$.
Q2 in $X = \mathbb R^d$: Suppose that $y := (0,1)$ is not an interior point of $K$(!). In case that the interior of $K$ is empty $K$ is contained in a proper subspace and we can take a $w$ which is orthogonal to this subspace (maybe we have to multiply by $-1$ to achieve $w_2 \ge 0$).
In case that the interior of $K$ is nonempty, we can use a separation theorem to separate the interior of $K$ and $(0,1)$. Thus, we obtain a non-zero $w$ and some $c$ with $$ \langle (0,1),w\rangle \ge c \ge \langle z,w\rangle $$ for all $z \in K$. Since $K$ is a cone we get $c = 0$ and the left-hand side yields $w_2 \ge 0$.