Nuclear spaces and intuition behind their topology

Question

In functional analysis the nuclear spaces (coined by Grothendieck before he became involved in revolutionizing algebraic geometry) can be considered as a kind of generalization of finite dimensional Euclidean spaces. While the Banach spaces represent another generalization of finite-dimensional vector spaces focused on retaining the properties of norm (and completeness with respect to it) but losing compactness properties, the focus of the nuclear spaces lies on retaining compactness properties.

In Wikipedia is remarked that informally one can think about the nuclear spaces to be characterized by the property that their topology is defined by a family of seminorms whose unit balls "decrease rapidly in size".

In more detail this means that whenever we are given the unit ball $B_{1,p}(0) \subset X$ with respect some seminorm $p: X \to \mathbb{R}$, we can find a "much smaller" unit ball $B_{1,q}(0) \subset X$ with respect another seminorm $q: X \to \mathbb{R}$ inside it, or that any neighborhood of $0$ contains a "much smaller" neighborhood.

What does it mean precisely that the unit ball $B_{1,q}(0)$ is "much smaller" than the unit ball $B_{1,p}(0)$? "Much smaller" in what sense? Can it be made precise what this means here? (the question is really about the intuition)

Answer 1

Some night thoughts on this question. At the level of operators, bounded operators are analogous to bounded sequences, compact to $c_0$ ones and nuclear to $\ell^1$. This analogy is precise in the sense that if we regard a sequence as a multiplication operator $T$ on $\ell^2$ in the usual way, then the bounded ones correspond exactly to $\ell^\infty$ and so on.

For nuclear spaces, we remain for the moment in the context of sequence spaces. Suppose that we have an unbounded sequence $(\lambda_n)$ with terms $\geq 1$. Then we can create, in a simple and natural way, a sequence $H^n$ ($n$ a whole number) of Hilbert spaces, where $H^n$ is the domain of definition of $T^n$ for $n$ positive, and $H^{-n}$ is its dual (under the scalar product of $\ell^2$). Then we can close the sequence with the intersection $\bigcap H^n$ and the union $\bigcup H^{-n}$ and these are an $F$-space and a $DF$-space respectively. (The Hilbert spaces can easily be identified with weighted $\ell^2$´s).

The connection with nuclearity lies in the fact that the limit spaces are nuclear when $\left(\frac 1{\lambda_n}\right) $ is in $\ell^1$.

The connection with the real world (i.e., the classical nuclear spaces of test functions and distributions) arises from the fact that we can, using the spectral theorem, carry out the same construction with an unbounded self-adjoint operator $T$ on Hilbert space. The resulting limit spaces are again nuclear, if $T$ has a discrete spectrum with eigenvalues $(\lambda_n)$ which satisfy the above condition. (In fact it suffices that there exist a positive $\alpha$ so that $\left(\frac 1{\lambda_n^\alpha}\right)$ in $\ell^1$).

If one applies this construction to classical operators (Sturm-Liouville, Laplace, Schrödinger) whose spectral properties are known, then one obtains a unified approach to many of the classical spaces of distribution theory.