Let $K$ be a closed convex cone. Then $K$ is solid if and only if $K$ is reproducing.
Hint:
- If $K$ is a convex cone then $K-K$ has a nonempty interior.
- $K-K$ is the minimal subspace containing $K$.
Definitions:
- A set $K$ in a Euclidean space $V$ is a convex cone if for every $x, y \in K$ and nonnegative scalars $\alpha, \beta$, $$\alpha x+\beta y\in K.$$
- A convex cone $К$ in $V$ is pointed if $K \cap (-K) = \{0\}$, reproducing if $K-K=V$, solid if $\operatorname{int} K\neq \emptyset$,
Assume $K$ is solid. Then it contains a ball $B_r(x)$. This implies $K-K$ contains $B_r(x)-B_r(x)$, which contains $B_r(0)$. Now $K-K$ is also a cone, hence $K-K=V$.
The reverse direction is not valid in general (think non-negative functions in $L^2$), so it is going to be ugly.
Assume $K-K=V$. Let $\dim V=n<+\infty$. Then for all unit vectors $e_i$ there are $x_i,y_i\in K$ such that $e_i = x_i-y_i$. It follows that $V$ is the conical hull of the $2n$ vectors $\{x_i,-y_i: i=1\dots n\}$. By the conical version of Caratheodory theorem, we can select at most $n$ vectors $v_j \in \{x_i,-y_i: i=1\dots n\}$, $j=1\dots m$, $m\le n$, such that $V$ is the conical hull of these $v_j$. It follows $m=n$. This implies that $K$ contains $n$ linearly independent vectors. And $K$ has non-empty interior: Since $\dim(span(K))=n$ it follows $int K = rel\ int (K)$, which is non-empty.