# Floer's space closed under products?

Floer (in "The unregularized gradient flow of the symplectic action") defined a dense subspace of $C^\infty$ with the structure of a Banach space, with norm $\Vert f \Vert = \sum_{k \ge 0} \epsilon_k \Vert f \Vert_{C^k}$ for some constants $\epsilon_k$.

Question: How do the constants $\epsilon_k$ in Floer's norm behave as $k \to \infty$? He just shows that some constants exist for which the resulting space is dense in $C^\infty$. Do the constants go to infinity, to zero, or neither?

Subquestion: Is Floer's space closed under products of functions?

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In case $\epsilon_k\sim (k!)^s$ this essentially gives Gevrey classes, see my other answer. More general (log-convex) sequences give rise to the so called Mandelbrojt classes. Not that Mandelbrojt, an older one. – Piero D'Ancona Aug 20 '10 at 15:25
er.. I mean $\epsilon_k \sim (k!)^{-s}$ – Piero D'Ancona Aug 20 '10 at 15:37

Floer chooses $\epsilon_k$ so that this $\epsilon$ space is dense in $L^2$ (this should be equivalent to dense in $C^\infty$, since the latter is dense in $L^2$, and Floer's $\epsilon$ space sits in $C^\infty$).

To show that this subspace is dense in $L^2$, it suffices to show that one can approximate indicator functions.
For this, he needs the $\epsilon_k$ to go to zero very fast. His explicit construction in Lemma 5.1 (of the "unregularized gradient flow" paper) is to take a fixed cut-off function $\beta$, and approximate the characteristic function of a rectangle. The approximation to a characteristic function is going to be a product of terms that behave like $\beta(x/\delta)$ (with a better approximation as $\delta \rightarrow 0$). Thus, the behaviour of the $\epsilon$-norm is going to be roughly: $\sum_{k=0}^\infty \epsilon_k \delta^{-k} a_k,$ where $a_k = \sup | D^k \beta |$. We need this to converge for each $\delta > 0$. Floer takes $\epsilon_k = (a_k k^k)^{-1}$. In particular, then, these constants are going to $0$. Following this argument, it seems we can take the $\epsilon_k$ to be on the order of $1/k!$.

Note that if the sequence $\epsilon_k$ is not summable, we expect the space to be very small. In particular, consider this norm on a compact interval, say $[-\pi, \pi]$. Then, cos(x) is not in the space.

The Floer $\epsilon$ space forms a Banach algebra if $\epsilon_k$ decays faster than $1/k!$.
Then, $\sum \epsilon_k |D^{(k)}(fg)| \le \sum_{k=0}^\infty \sum_{l=0}^{k} \epsilon_k | D^{(l)} f| |D^{(k-l)} g | \binom{k}{l} = \sum_{l=0}^\infty |D^{(l)} f| \sum_{p=0}^{\infty} \binom{l+p}{p} \epsilon_{p+l} | D^{(p)}g|$ When $\epsilon_{p+l} \binom{l+p}{p} \le \epsilon_p \epsilon_l$, we are then in business. In particular, this works for Floer's original construction.

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Sam, do you have a lower bound on $a_k$? – Tim Perutz Nov 3 '10 at 14:54
If we follow Floer's definition of the cut-off functions, we can crudely bound them below by 1. To recall, Floer's definition is that $\beta : \mathbb{R} \rightarrow [0,1]$ with $\beta(x) = 0$ near $0$ and $\beta(x) = 1$ near 1, $\beta'(x) \ge 0$. He then uses products of functions of the form $\beta(|x-y|/\delta)$ to approximate characteristic functions of rectangles. Actually, we might be able to do better and get something like $1/d^k$, where d is the diameter of the support of $\beta$. (The exact bound doesn't really matter because of the $1/\delta^k$ term that shows up.) – Sam Lisi Nov 3 '10 at 23:18
Ah yes - by the mean-value theorem and induction. (For the uniform lower bound, I mean.) Missed that. Thanks! – Tim Perutz Nov 4 '10 at 13:46