If $\Bbb C^\infty$ is the only infinite-dimensional Hilbert space, why do other space names need to be used?

Question

In Dym and McKean book, the map ^ originally start out as an isomorphism $L^2(Q)$ ($Q$ is any finite measure interval) to $L^2(\mathbb Z^+)$ (which is just another name of $\mathbb{C}^\infty$), therefore the coefficients are summed from $1$ to $\infty$. As I understand here it comes from the definition of Lebesgue integral. Later, when $Q=S^1$, it maps to $L^2(\mathbb Z^1)$, making the summation ranges from $-\infty$ to $\infty$. The problem is

Up to isomorphism, $\Bbb C^\infty$ is the only infinite-dimentional Hilbert space there is.

If so, why do we need to separate $L^2(\mathbb Z^+)$ and $L^2(\mathbb Z^1)$?

score 1 · Accepted Answer · answered Nov 23 '17 at 00:34

If you start with a concrete mathematical problem, such as

for given function $p,q$ on an interval $I$, find functions $f$ that satisfiy $\lambda f = -(p(f')' + qf$

then you start with a function space and you can ask this question for $f \in L^2(I)$ and end up with a differential operator (in this case a Sturm-Liouville operator) in $L^2(I)$. You can write down what the operator does in that setting. However, if you use the isomorphism which maps $L^2(I)$ to $\mathbb C ^\infty$, you need to transform the operator and the pre-images as well and you lose easy access to some information:

What's the action of the differential operator?
We can't integrate by parts to manipulate algebraic statements depending on the operator
How does a boundary value of a function translate to $\mathbb C ^\infty$
Given two functions, how is the scalar product translated?

To wrap up, different Hilbert spaces are driven by applications they are used in and some properties of elements of these are possily easier worked with than an element the image of an isomorphism.

Is there a case that using the isomorphism is preferable? Also, do you know what the action of the differential operator is in $\Bbb C^\infty$? — Ooker, Nov 23 '17 at 00:40
Follow up question: How is orthonormal basis ${e^{2πikt}}_{k=-\infty}^\infty$ with index going to $-\infty$ constructed in $L^2(\Bbb S^1)$? — Ooker, Nov 27 '17 at 05:54

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

I think it's just the choice of convenience, since $\Bbb Z^+$ has a bijection to $\Bbb Z$. From comments:

We can index every countably infinite family by $\mathbb{Z}^+$. But in $L^2(S^1)$ (with the measure normalised) we have the useful orthonormal basis $\{e_n \colon z \mapsto z^n \mid n \in \mathbb{Z}\}$ which is much more convenient to index by $\mathbb{Z}$ rather than $\mathbb{Z}^+$.

For particular measure spaces, there may be particularly obvious or useful orthonormal bases, but in general, an orthonormal basis can be hard to find.

http://people.csail.mit.edu/alinush/math/countability.pdf

If $\Bbb C^\infty$ is the only infinite-dimensional Hilbert space, why do other space names need to be used?

2 Answers2