Let $f$ be a bounded holomorphic function on $\mathbb D^2$ and $s : \mathbb C^2 \longrightarrow \mathbb C^2$ be the symmetrization map given by $s(z) = (z_1 + z_2, z_1 z_2),$ for $z = (z_1, z_2) \in \mathbb C^2.$ If $\{(f \circ s) g_n\}_{n \geq 1}$ has a convergent subsequence in $L^2 (\mathbb D^2)$ for any sequence of bounded anti-symmetric holomorphic square integrable functions $\{g_n\}_{n \geq 1}$ then what can we conclude when $g_n$'s are replaced by a sequence of bounded symmetric square integrable holomorphic functions?
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- $\begingroup$ Please define symmetric and anti-symmetric functions. $\endgroup$Vít Tuček– Vít Tuček2024-06-22 21:45:31 +00:00Commented Jun 22, 2024 at 21:45
- $\begingroup$ @VítTuček$:$ Symmetric functions are those which are invariant under the $S_2$-action on $\mathbb A^2 \left (\mathbb D^2 \right)$ and anti-symmetric functions are those that are altered according to the permutations of $S_2.$ $\endgroup$Anacardium– Anacardium2024-06-23 06:05:40 +00:00Commented Jun 23, 2024 at 6:05
- $\begingroup$ So anti-symmetric functions are all those which are not symmetric? $\endgroup$Vít Tuček– Vít Tuček2024-06-23 22:40:47 +00:00Commented Jun 23, 2024 at 22:40
- $\begingroup$ @VítTuček$:$ Yes. More precisely $f$ Is anti-symmetric on $\mathbb D^2$ if $f(z_1,z_2) = - f(z_2,z_1),$ for all $(z_1,z_2) \in \mathbb D^2.$ $\endgroup$Anacardium– Anacardium2024-06-24 05:05:15 +00:00Commented Jun 24, 2024 at 5:05
- $\begingroup$ So, strictly speaking, no anti-symmetric functions are not the set theoretic complement of the symmetric functions. :) But I think that there is orthogonal decomposition of $L^2(\mathbb{D})$ into symmetric and anti-symmetric functions (sending $f$ to the pair $(1/2[f(a,b) + f(b,a)], 1/2[f(a,b)-f(b,a)]$). Is that your setting? $\endgroup$Vít Tuček– Vít Tuček2024-06-24 18:17:23 +00:00Commented Jun 24, 2024 at 18:17
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