Timeline for Orthonormal bases on Reproducing Kernel Hilbert Spaces
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2016 at 12:58 | comment | added | T. Le | Update on June 8, 2016: I saw in a recent book "An Introduction to the Theory of Reproducing Kernel Hilbert Spaces" by V. Paulsen and M. Raghupathi, that the proof of Moore-Aronszajn (Theorem 2.14 on page 25) has been done more carefully. In particular, the injectivity was carefully checked. | |
| Jan 24, 2016 at 3:23 | vote | accept | T. Le | ||
| Jan 24, 2016 at 3:06 | answer | added | Nate Eldredge | timeline score: 8 | |
| Jan 24, 2016 at 1:13 | comment | added | T. Le | @NateEldredge: I think you are right! I was unable to verify that the map from $\widetilde{\mathcal{M}}$ to $\mathbb{R}^{X}$ (or $\mathbb{C}^{X}$ for the complex case) is injective. In other words, if $h\in\widetilde{\mathcal{M}}$ and $h(x)=0$ for all $x\in X$, then it cannot be verified that $\|h\|=0$ (in fact, this may not be true). On the other hand, in the "standard proofs" of Moore-Aronszajn Theorem, injectivity can be verified (even though in all the proofs that I know of, it was not mentioned or verified!!). Could you please put part of your comments in an answer so I can vote for it? | |
| Jan 23, 2016 at 23:52 | comment | added | Nate Eldredge | In other words, could there be two inequivalent norm-Cauchy sequences $\{f_n\}, \{g_n\}$ (i.e. $\|f_n - g_n\| \not\to 0$) which have the same pointwise limit $f$? If so, then you can't view $\widetilde{\mathcal{M}}$ as a space of functions, since it would be ambiguous whether the function $f$ represents the element $[\{f_n\}]$ or $[\{g_n\}]$ of $\widetilde{\mathcal{M}}$. | |
| Jan 23, 2016 at 23:50 | comment | added | Nate Eldredge | So we can map the equivalence class $[\{f_n\}]$ to the function $f : X \to \mathbb{R}$ which is the pointwise limit of the norm-Cauchy sequence $\{f_n\}$, and it is easy to check that this is well-defined, i.e. two equivalent Cauchy sequences get mapped to the same function. So we have a well-defined linear map from $\widetilde{\mathcal{M}}$ to $\mathbb{R}^X$. But in order to be able to view $\widetilde{\mathcal{M}}$ as a space of functions, we need this map to be injective, and that part is not clear to me. (continued) | |
| Jan 23, 2016 at 23:46 | comment | added | Nate Eldredge | A priori, an element of the completion $\widetilde{\mathcal{M}}$ is not a function on $X$, but an equivalence class of norm-Cauchy sequences of functions. To be able to consider $\widetilde{\mathcal{M}}$ as an RHKS, which must be a space of functions, we must identify each equivalence class $[\{f_n\}]$ with an actual function on $X$. Now since the evaluation functional is (uniformly) continuous on $\mathcal{M}$, it is clear that any norm-Cauchy sequence is also pointwise Cauchy, hence pointwise convergent. (continued) | |
| Jan 23, 2016 at 23:02 | comment | added | T. Le | By "standard argument", I meant the usual argument that one might find when proving that any positive definite kernel K(x,y) gives rise to an RKHS whose kernel is K(x,y) (Moore-Aronszajn Theorem). Such argument can be found in Vern Paulsen's note, page 14, at www.math.uh.edu/~vern/rkhs.pdf or in the proof of Theorem 2.23 on page 19 in Agler-McCarthy's book "Pick Interpolation and Hilbert Function Spaces". @NateEldredge: can you explain what you mean and why you need the association to be 1-1 for $\widetilde{\mathcal{M}}$ to be RKHS? | |
| Jan 23, 2016 at 20:38 | comment | added | Nate Eldredge | This sentence worries me: "The standard argument shows that $\widetilde{\mathcal{M}}$ is a RKHS of functions on $X$". Can you look at that more carefully? I am not convinced that the completion $\widetilde{\mathcal{M}}$ can be identified with a space of functions on $X$. You have shown that the map $f \mapsto f(x)$ is continuous with respect to the $\mathcal{M}$ norm, so this map extends to the completion, and every element $\tilde{f}$ of the completion is associated with a well-defined function on $X$. But I do not see why this association has to be 1-1. | |
| Jan 23, 2016 at 19:59 | comment | added | T. Le | @ChrisRamsey: Together with Theorem 3.12 in Vern Paulsen's lecture note at www.math.uh.edu/~vern/rkhs.pdf, if the conclusion were correct, then any linearly independent Parseval frame would be an orthonormal basis. However this seems to be incorrect. For example, take E to be the orthogonal complement of v = (1, 1/2, 1/3, ...) in l^2 and take g_m = P(e_m), where P is the orthogonal projection onto E. Then {g_m} is a linearly independent Parseval frame for E but it is not an orthonormal basis. | |
| Jan 23, 2016 at 19:43 | comment | added | Chris Ramsey | Could you describe briefly one of these counterexamples? | |
| Jan 23, 2016 at 19:32 | review | First posts | |||
| Jan 23, 2016 at 21:54 | |||||
| Jan 23, 2016 at 19:27 | history | asked | T. Le | CC BY-SA 3.0 |