I'm not familiar with the infinite-dimensional case. Do you have an example of a tensor that isn't a multilinear function on a product of vector spaces? I'd be interested to refine my understanding here.
A familiar example of a tensor product of countably infinite dimensional vector spaces would be polynomials in multiple variables. Say F[x] is the vector space of polynomials in x with coefficients in the field F, and F[x,y] is the space of polynomials in the variables x, y. For example x^2 - x is an element of F[x], and yx^2 - y^2 is an element of F[x,y]. Then it’s not hard to see that as vector spaces, F[x,y] is isomorphic to (F[x] tensor F[y]). A tensor in this new space is precisely a polynomial in two variables.
In the above it’s important to note that a polynomial has finitely many terms, so a power series like 1 + x + x^2 + … is not a polynomial. The space of power series is isomorphic to the space of linear functions on F[x], and is uncountably-infinite dimensional.
I think I'm missing part of the argument. So we have a polynomial p in F[x,y]. The claim in my previous post basically says that there's always a way to associate p with a linear map that takes two elements of the dual space of F[x] and produces a real number. I don't see why that's impossible here.
So every polynomial could be represented as a linear function on polynomials, but not every linear function on polynomials is itself a polynomial.
So we're talking about the case when tensors don’t need to be multilinear functions on a product of vector spaces.
Each element of V* is (trivially) a tensor and a linear function on V. Each element of V is (trivially) a tensor and a linear function on V*. However, not all linear functions on V* are in V. So V** is bigger than V. No problem so far.
But all elements of V** are functions on (trivial) products of vector spaces, and by definition all functions in V** are linear. So how have we misunderstood each other here?
In order to have V* = V for an infinite-dimensional vector space V, you need to redefine V^* to some kind of restricted dual, rather than defining it as the set of all linear functions. In the polynomial example, if we take the space of all linear maps g: F[x] -> F such that g(x^n) = g(x^(n+1)) = ... = 0 for some n >> 0, then this restricted dual is isomorphic to F[x] again. But the evaluation map g(f) = f(1) is not in this restricted dual.
There are more reasons why confusing a vector space with its dual is a bad idea. For example you cannot cook up a map V -> V* without extra knowledge, for example a choice of basis of V or something. There are many examples in abstract algebra where there is a perfectly good vector space V, and absolutely no good choice of basis for V, so trying to identify elements of V with V* is unnatural. We may still be able to speak perfectly well of vectors in V or V*, but trying to identify V with V* is still unnatural. A good example is V = (functions R -> R). I can speak easily of elements of V (for example, x + sin(x)), and of elements of V* (for example, f -> integral of xf(x)), but trying to figure out which element in the dual either of these corresponds to is hopeless. We're better off just accepting at some point that there is a real difference between a vector space and its dual.