# How is Schwartz space different from Hilbert space?

I know that Schwartz space can be considered a dense subset of the Hilbert space isomorphic to $$\ell^2$$. What I wish to understand is, how really different Schwartz space is from the Hilbert space.

Schwartz space has completeness, and is an inner product space, at least understood as "borrowing" inner product structure from the Hilbert space. This seems to satisfy all the definitions of the Hilbert space. Yet we know that these two spaces are not isomorphic. So what distinguish them?

Maybe the problem is that inner product borrowed from the Hilbert space is not natural to the natural topology of Schwartz space? It would be nice if someone clarifies on this.

And I believe quantum mechanics uses inner product on Schwartz space, so I guess Schwartz space is understood at least there as an inner product space...

• What is the inner product on the Schwartz space? Usually the Schwartz space is given a family seminorms. Mar 10 '19 at 0:24
• @rolandcyp That, I would have to leave empty, as the question is asked out of confusion. Maybe one can start from the inner product borrowed from Hilbert space, but unnaturalness of that inner product may be why these Schwartz space and Hilbert space are not really isomorphic. Again, I am in state of confusion, so there may be some confusion-clearing needed. Mar 10 '19 at 0:40
• Which Hilbert space do you have in mind specifically? Given that we are discussing $\mathscr{S}$, I assume you mean $L^2$? Mar 10 '19 at 0:42
• @rolandcyp Yes. That's correct. Mar 10 '19 at 0:43
• Sure there is an inner product. But the Schwartz space is not complete for the norm defined by the inner product. This subspace of Hilbert space is not a closed subspace in the Hilbert space norm. Mar 10 '19 at 0:45

Let $$\mathscr{S}$$ denote the Schwartz space. We equip this space with a topology by giving it the family of seminorms $$p_{\alpha, \beta}(f) := \sup_{x \in \mathbb{R}^n} |x^\beta \partial^\alpha f(x)|$$ where $$\alpha,\beta$$ are multi indices. With respect to the induced topology, we know that $$\mathscr{S}$$ is completely metrizable, as you have noted.
On the other hand, one can view $$\mathscr{S}$$ as a subspace of $$L^2(\mathbb{R}^n)$$. Hence, $$\mathscr{S}$$ inherits an inner-product (and a metric topology). However, this is not the same topology as above! Indeed, $$\mathscr{S}$$ is not even complete with respect to the metric inherited from $$L^2(\mathbb{R}^n)$$. This is because $$\mathscr{S}$$ is not closed as a subspace of $$L^2(\mathbb{R}^n)$$. To see this, simply note that $$\overline{\mathscr{S}} = L^2(\mathbb{R}^n) \neq \mathscr{S}$$ when interpreted as a subspace of $$L^2(\mathbb{R}^n)$$.
• And $\mathscr{S}$ is the intersection of all the Hilbert spaces $H^k$ and their Fourier transforms but there is no inner product making it an Hilbert space (proof ?) we can only look at metrics like $\sum_{k\ge 1} 2^{-k}\frac{\|f\|_{H^K}}{1+\|f\|_{H^K}}+2^{-k}\frac{\|\hat{f}\|_{H^K}}{1+\|\hat{f}\|_{H^K}}$ Mar 10 '19 at 8:17
• How does incompleteness imply that the inner product topology not coincide with the standard topology? You can just say that it is strictly finer. For example, the set of functions with $\lvert f'\rvert<1$ is obviously not open in the inner product topology. Mar 13 '19 at 16:19
• And I assume that there's no way to assign a different inner product to $\mathcal{S}$ that is compatible with the former topology? Apr 10 at 16:38
Since you mentioned $$\ell^2$$ and quantum mechanics, here is a clean way to see the difference between the Schwartz space $$\mathscr{S}(\mathbb{R})$$ and $$L^2(\mathbb{R})$$. Take the basis of Hermite functions $$(\psi_n)_{n\ge 0}$$ which diagonalizes the Hamiltonian for the harmonic oscillator $$H=\frac{1}{2}(P^2+X^2)$$. This gives an isomorphism $$T:L^2(\mathbb{R})\rightarrow \ell^2$$, $$f\rightarrow \langle \psi_n,f\rangle_{L^2}$$. The image of $$\mathscr{S}(\mathbb{R})\subset L^2(\mathbb{R})$$ is the space $$\mathfrak{s}$$ of sequences $$x=(x_n)_{n\ge 0}$$ with rapid decay. These are the sequences such that for all $$k\ge 0$$, $$||x||_k=\sup_{n\ge 0}\ (n+1)^k|x_n|$$ is finite. In fact, the map $$T$$ realizes a topological vector space isomorphism between $$\mathscr{S}(\mathbb{R})$$ and $$\mathfrak{s}$$ if the latter is equipped with the topology defined by the seminorms $$||\cdot||_{k}$$, $$k\ge 0$$.