Timeline for Orthonormal bases on Reproducing Kernel Hilbert Spaces
Current License: CC BY-SA 3.0
7 events
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| Jan 24, 2016 at 3:59 | comment | added | Nate Eldredge | @T.Le: Yes, that's true, because the map $S : \ell^2 \to \widetilde{\mathcal{M}}$ which maps $\{c_m\}$ to $ \sum_m c_m g_m$ is an isometric isomorphism. Your condition is that $TS$ is injective, which means that $T$ must be as well. | |
| Jan 24, 2016 at 3:46 | comment | added | T. Le | Thank you for your answer! On the other hand, I believe that the map $T$ is injective and hence the Question has an affirmative answer if we assume a stronger condition than just linear independence: for any sequence $\{c_m\}$ in $\ell^2$, if $\sum_{m=1}^{\infty}c_mg_m(x) = 0$ for all $x\in X$, then $c_1=c_2=\cdots = 0$ (note that the convergence of the series is automatic by Cauchy-Schwarz's inequality). I think there is a name for this condition but I'm not sure what it is. | |
| Jan 24, 2016 at 3:34 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
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| Jan 24, 2016 at 3:25 | comment | added | Yemon Choi | Nice answer. Just a remark related to your last point: we get particularly striking examples if X is the full group Cstar algebra of a non-amenable discrete group (like $F_2$), E is the dense subalgebra generated by point masses, and Y is the reduced group Cstar algebra. Then the "identity map" from E to Y is continuous and injective with dense range, but the unique continuous extension to a map $X \to Y$ is not injective | |
| Jan 24, 2016 at 3:23 | vote | accept | T. Le | ||
| Jan 24, 2016 at 3:21 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
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| Jan 24, 2016 at 3:06 | history | answered | Nate Eldredge | CC BY-SA 3.0 |