Essentially, the totally convex spaces are what you get if you try to build an algebraic theory of Banach spaces.
A totally convex space is a module for the monad on Set which sends a set $X$ to the closed unit ball of the Banach space $\ell^1(X)$.
This is an example of an algebraic theory (in the strong sense that it is bounded monadic) but not finitary, hence not a Lawvere theory. The free totally convex space on $X$ is the unit ball of the Banach space $\ell^1(X)$, and thus an operation in this theory is a formal sum $\sum_{x \in X} a_x x$ for $a_x \in \mathbb{R}$ with the property that $\sum_{x \in X} {|a_x|} \le 1$. The finiteness of this sum forces it to have only countably many non-zero terms, and thus it factors through an operation on $\mathbb{N}$. Hence there is a presentation of this theory with operations in $\ell^1(\mathbb{N})$ (which is why the theory is bounded monadic). The corresponding identities are simply the substitution rules – namely that substituting a sum into another works as expected – and the reordering rules.
A totally convex space is a set $X$ equipped with, for each infinite sequence $(a_1, a_2, \ldots)$ of real numbers such that $\sum_i {|a_i|} \leq 1$, an operation from $X^{\mathbb{N}}$ to $X$, written $(x_1,x_2,\ldots) \mapsto \sum_i a_i x_i$, such that:
Of course, one would normally write the right-hand side of the last equation as $\sum_{i,j} a_i b_{i,j} x_{i,j}$, but that is not technically an operation in the theory, except as mediated by $\pi$ and $\rho$. A common choice for $(\pi,\rho)$ is the inverse of $(i,j) \mapsto \big({i + j + 1 \atop 2}\big) + j$, where for this expression to work we take $0$ to be a natural number.
I haven't actually checked that this list is complete; but it's what I get if I take Andrew at his word that we need only substitution and reordering rules. I wouldn't be terribly surprised if nullary substitution is redundant, but right now I don't see how. —Toby
A totally convex space is a pointed convex space. The operations of a convex space are encoded in the operations $(r, 1 - r, 0, 0, 0, \ldots)$ and the “point” comes from the operation $(0, 0, \ldots)$. This functor preserves underlying sets and so has a left adjoint; thus any convex space can be “completed” to a totally convex space.
The operations for this theory are commutative, hence the category of totally convex spaces is a closed symmetric monoidal category.
Clearly, the closed unit ball $B X$ of any Banach space $X$ is a totally convex space.
Bizarrely, the open unit ball of a Banach space is a totally convex space. This is because if a sum, $\sum_{x \in B X} a_x x$ for $\sum_x {|a_x|} \leq 1$, lies on the boundary of $B X$ then every $x$ for which $a_x \ne 0$ must have norm $1$. Thus if the series only contains terms from the interior of $B X$, the sum remains in the interior. Hence the open unit ball is a subalgebra of the closed unit ball.
Continuing, the quotient of the closed unit ball by the open unit ball is a totally convex space.
In particular, the three-point space $\{-1,0,1\}$ is (assuming excluded middle) a totally convex space. The operations on this space are as follows:
Going back one step, $(-1,1)$ is a totally convex space. It is illuminating to describe this as a coequaliser of free totally convex spaces. Consider a functional $f \colon \ell^1(\mathbb{N}) \to \mathbb{R}$ which is bounded of norm $1$ but does not achieve its norm; for example, let $f$ be represented by the sequence $(\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots)$. Then for any $x \in B\ell^1(\mathbb{N})$, $f(x) \in (-1,1)$. Thus $(-1,1)$ is the coequaliser of the inclusion $\ker f \to \ell^1(\mathbb{N})$ and the zero map $\ker f \to \ell^1(\mathbb{N})$.