Floer's space closed under products? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:53:21Z http://mathoverflow.net/feeds/question/36192 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36192/floers-space-closed-under-products Floer's space closed under products? Chris Woodward 2010-08-20T14:34:48Z 2010-11-03T05:42:56Z <p>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 $\epsilon_k$.</p> <p>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? </p> <p>Subquestion: Is Floer's space closed under products of functions? </p> http://mathoverflow.net/questions/36192/floers-space-closed-under-products/44650#44650 Answer by Sam Lisi for Floer's space closed under products? Sam Lisi 2010-11-03T05:42:56Z 2010-11-03T05:42:56Z <p>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 $C^\infty$). </p> <p>To show that this subspace is dense in $L^2$, it suffices to show that one can approximate indicator functions.<br> 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 $1/k!$.</p> <p>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. </p> <p>The Floer $\epsilon$ space forms a Banach algebra if $\epsilon_k$ decays faster than $1/k!$.<br> 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 <code>$\epsilon_{p+l} \binom{l+p}{p} \le \epsilon_p \epsilon_l$</code>, we are then in business. In particular, this works for Floer's original construction.</p>