Banach Spaces, Inseparable Hilbert Space

Countable Representation

If b is an orthonormal system in a possibly inseparable hilbert space S, then any vector v has countably many nonzero coefficients with respect to b.

Suppose v has uncountably many nonzero coefficients in its representation. Let ℵ₁ be the first uncountable ordinal. Assign an ordinal below ℵ₁ to each of these nonzero coefficients. In other words, put them in order.

Consider a finite sum y = a₁b₁+a₂b₂+a₃b₃. Let z = x-y. Now z.b₁ = a₁-a₁ = 0. By linearity, z.y = 0. Expand (y+z).(y+z) and get |y|²+|z|². This equals |x|². Thus a₁²+a₂²+a₃² ≤ |x|². The same reasoning holds for any finite collection of terms drawn from x.

Let d be an ordinal below ℵ₁. Consider all the finite sets of coefficients with ordinals at or below d. For each finite set, the sum of squares is bounded by |x|². Let e_d be the supremum of these real numbers. Of course e_d is vounded by |x|².

Advance to d+1, the successor of d. Finite sums come arbitrarily close to e_d. Bring in the next coefficient, a_d+1², and that pushes some of these finite sums over the top, beyond e_d. Thus e_d+1 > e_d. There is a rational number between e_d and e_d+1. This holds for every ordinal d, hence there are uncountably many rational numbers between 0 and |x|², which is a contradiction.

Given v, restrict attention to those basis elements with v.b_i nonzero. As before, the coefficients a_i form a square summable sequence, the partial sums over a_ib_i are cauchy, the limit u exists in the complete metric space S, u and v produce the same coefficients, and v-u is orthogonal to each b_j. If b is a complete orthonormal system, then v-u = 0, and v is faithfully represented by a countable set of nonzero coefficients, applied to the basis b.

Bounded Linear Operator is Dot Product

Let f be a bounded linear operator from S into R. As you recall, c_i = f(b_i), and each c_i is bounded by k. Suppose uncountably many basis elements b_i have nonzero images in R. Associate these images with the ordinals below ℵ₁. For any ordinal d, let e_d be the supremum of ∑ c_i², over all finite sets drawn from the basis elements at or below d. An argument identical to the one presented above builds an uncountable increasing sequence of rational numbers between 0 and k, which is a contradiction. Thus f(b_i) is zero most of the time.

Concentrate on the countable subsequence b_i, where c_i is nonzero. Let u be the infinite (or perhaps finite) sum over c_ib_i. The earlier proof applies. A linear bounded/continuous operator is equivalent to S.u, and the bound is |u|, realized by f(u).

The Invariant Dimension Property

If the cardinality of b is g, then we need at least g elements in any dense set. This is because the elements of b are all sqrt(2) units apart, and can be enclosed in disjoint open balls of radius ¼. Conversely, we can produce a dense set with g elements by taking all finite linear combinations of b with rational coefficients. Each v is a countable linear combination drawn from b, and the finite linear combinations approach v. Find a finite partial sum within ½ε of v, then adjust the coefficients to nearby rational numbers, so that the adjustment doesn't stray by more than ½ε. The result is inside the open ball about v with radius ε.

Beyond Rⁿ, the dimension of S, as a hilbert space, is equal to the size of the smallest dense set in S. The latter is a function of the topology. Thus the dimension of a hilbert space is well defined. If you want to change the dimension, you have to change the topology of S, or find a new set altogether.

Two hilbert spaces are isomorphic iff they have the same dimension. The hilbert space of dimension g has an orthonormal basis of size g, and all countable linear combinations thereof, such that the coefficients are square summable. To prove this construction is in fact a hilbert space, review the earlier theorem for a separable space. Not much has changed, as long as you remember that the union of countable sets remains countable. The sum of two elements in S still draws on a countable subset of b. Even the infinite countable union of countable sets is countable. This is used to prove S is complete. The cauchy sequence is countable, and each element in this sequence uses a countable subset of b, hence the entire cauchy sequence lives in a seprable hilbert subspace of S, and approaches its limit.

In summary, there is one hilbert space, up to isomorphism, for each nonzero cardinal, and the space is completely characterized as the countable linear combinations of basis elements, with square summable coefficients.