How can tempered distributions be paths? I'm reading the Appendix A of Glimm and Jaffe book "Quantum Physics: a functional integral point of view", and there is something that I'm missing
In section A.4 the authors talk in a very general context about functional integration. If I got it right they are considering a sequence of Hilbert spaces $\mathscr{H}_n$ and setting $$\mathscr{H}_\infty=\bigcap_{n\in \mathbb{Z}}\mathscr{H}_n,\quad \mathscr{H}_{-\infty}=\bigcup_{n\in \mathbb{Z}}\mathscr{H}_n$$
They say $\mathscr{H}_\infty$ is a nuclear space and $\mathscr{H}_{-\infty}$ its dual. They exemplify with $\mathscr{S}$ the Schwartz space and $\mathscr{S}'$ the corresponding distributions.
Then the authors set out to study measures and integration on $\mathscr{H}_{-\infty}$:

We take Gaussian measures as the starting point for integration over infinite dimensional spaces. Other, non-Gaussian, measures are then obtained by perturbation, e.g., through the Feynman-Kac formula. The dual of a nuclear space (i.e. $\mathscr{H}_{-\infty}$) provides a convenient framework for studying Gaussian measures over infinite dimensional spaces.

I'm really missing the point of considering this kind of Hilbert spaces, 
specially these ones defined by these sequences.
Further on the authors even call an element of $\mathscr{S}'$ a path. How can that be? A path is a mapping $\gamma : [a,b]\to \mathbb{R}^d$ and an element of $\mathscr{S}'$ is a map $\varphi : \mathscr{S}\to \mathbb{R}$ acting on functions$f : \mathbb{R}^d\to \mathbb{R}$. I can't see why an element of that space is a path!
Also I never thought that the space of paths needed to carry any inner product structure. For instance, I always considered that the relevant space for non-relativistic quantum mechanics was $C^0([a,b];\mathbb{R}^d)$.
So what is the intuition here? Why consider Hilbert spaces - and hence an inner product structure - as the spaces of functions one is integrating over? Furthermore, why nuclear spaces?
 A: If you read through the book in chronological order, Glimm and Jaffe already have given you all the puzzle pieces. The first is Minlos' theorem (theorem 3.4.2 of their book), which says that a functional $S$ on smooth compactly supported functions on $\mathbb{R}^n$ defines a measure $\mathrm{d}\mu$ on the dual space of these functions, and that this measure is concentrated on the dual of the Schwartz functions if $S$ extends continuously to the Schwartz functions, and the functional and the measure are related by Fourier transformation:
$$ S[f] = \int \mathrm{e}^{\mathrm{i}\langle q,f\rangle}\mathrm{d}\mu(q)$$ 
Maybe the notation is throwing you off: Physicists would write $Z$ for $S$, $J$ for $f$ and would insist on expressing the measure by the "path-measure" $\mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q$, where $S$ is the ordinary classical action. That is, Minlos' theorem tells us where the individual components of the physical "path integral" have to live in order for the partition function $Z[J]$ to be well-defined mathematically. 
Then, in the very section you quote, they show you that a covariance operator on $H_\infty$ defines a Gaußian measure on its dual $H_{-\infty}$. The "covariance operator" $C$ applied to two $q$ is simply the quadratic part of the classical action, i.e. $\int \dot{q}(t)\dot{q}(t)\mathrm{d}t$ for a free particle.
This operator is certainly well-defined on Schwartz functions since the bound on their derivatives means this integral will exist. So the free particle action defines a covariance operator on the dual Schwartz space (the "paths"), hence a partition function in terms of Schwartz functions. Note that the action acts on Schwartz functions, but the integral is over the dual! So do we call the argument to the action "path" or the argument to the measure? Glimm and Jaffe arbitrarily decide to call the argument of the measure "path", but you may well decide differently - it is just a word, after all.
There is no "intuition" to this notion of "path". Glimm and Jaffe are not looking for intuition. Their goal is not to show an intuitive way of how one might have developed quantum physics, but of how one might put its well-known heuristic and hand-wavy formula on solid mathematical ground. The Hilbert space theory for the function spaces appears not because of any connection to quantum mechanical spaces of state but because the straightforward inner product $\langle f,g\rangle = \int f(x)g(x)\mathrm{d}^d x$ endows the $L^2$-space with an inner product structure that is useful because you can apply various theorems to it (such as the theorem about nuclear spaces and covariance operators above). The nuclear spaces appear simply because that's what you can prove the theorem for (nuclear spaces are "nice" infinite-dimensional spaces), not because of any physical intuition.
If you're looking for intuition, I guarantee you any standard physical treatment of path integrals will have more of it than this book.
