Timeline for Existence of a projection operator onto subspace of Hilbert space
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2016 at 12:09 | vote | accept | Charpe | ||
| Apr 1, 2015 at 21:02 | answer | added | James | timeline score: 2 | |
| Apr 1, 2015 at 21:01 | comment | added | Nate Eldredge | @ChristianRemling: Yes we can: if $v \in V_n^\perp$ (with respect to $H$-inner product) then for any $v_n \in V$ we have $$\langle Q_n v, v_n \rangle_H = \langle Q_n v - v, v_n \rangle_H + \langle v, v_n \rangle_H = 0+0$$ so that $Q_n v \in V_n^\perp$. Since $Q_n v \in V_n$ we have $Q_n v = 0$. | |
| Apr 1, 2015 at 20:01 | comment | added | Charpe | @NateEldredge Ok, I agree. | |
| Apr 1, 2015 at 19:49 | comment | added | Nate Eldredge | Of course, as I said, my new basis $\{u_j\}$ may not even be $V$-orthogonal, so I am not claiming this reduces the problem to the "standard" case. | |
| Apr 1, 2015 at 19:48 | comment | added | Nate Eldredge | @Charpe: I don't think you understand me. Even if $\{w_j\}$ is not orthogonal, I can use Gram-Schmidt to choose another set $\{u_j\}$ which is $H$-orthonormal and for which $V_n = \operatorname{span}\{w_1, \dots, w_n\} = \operatorname{span}\{u_1,\dots, u_n\}$ for every $n$. Your desired conditions only depend on the basis $\{w_j\}$ through the spaces $V_n$ which they define, so if I switch to a different basis for which the $V_n$ are the same, the truth of the conclusion does not change. There really is no loss of generality in that assumption. | |
| Apr 1, 2015 at 19:44 | comment | added | Nate Eldredge | @ChristianRemling: Isn't $Q_n$ forced to be a projection? If $v \in V_n$ then we have $Q_n v - v \in V_n$ and 1 says that $Q_n v - v \in V_n^\perp$. Hence $Q_n v = v$. | |
| Apr 1, 2015 at 19:43 | history | edited | Charpe | CC BY-SA 3.0 |
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| Apr 1, 2015 at 19:38 | comment | added | Charpe | I don't want to assume orthogonality, i.e., I want a proof without orthogonality because in some cases you want to use a very particular basis which is not orthogonal. (When the basis is orthogonal in $V$ and o.n in $H$, the standard proof of this theorem uses compactness of $V \subset H$ and Hilbert-Schmidt theory to get the desired result in the OP). | |
| Apr 1, 2015 at 19:37 | comment | added | Nate Eldredge | Also, I believe that thanks to Gram-Schmidt, we can assume without loss of generality that $\{w_j\}$ is $H$-orthonormal (or $V$-orthonormal, though of course not both). The required conditions only depend on the spaces $V_n$, not the particular basis chosen. | |
| Apr 1, 2015 at 19:33 | comment | added | Charpe | @NateEldredge Yes let's take separable spaces. If your second comment is right then maybe 2. cannot be proved abstractly. | |
| Apr 1, 2015 at 19:33 | history | edited | Charpe | CC BY-SA 3.0 |
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| Apr 1, 2015 at 19:32 | comment | added | Nate Eldredge | Unless I am missing something, doesn't 1 force $Q_n$ to be precisely the $H$-orthogonal projection onto $V_n$, restricted to $V$? | |
| Apr 1, 2015 at 19:21 | comment | added | Nate Eldredge | Is $\{w_j\}$ supposed to be countable? | |
| Apr 1, 2015 at 18:25 | review | Close votes | |||
| Apr 1, 2015 at 22:16 | |||||
| Apr 1, 2015 at 12:31 | comment | added | Charpe | @Hachino Note that 2. is asking for boundedness in $V$ to $V$, not in $H$ and 1. uses the inner product in $H$. | |
| Apr 1, 2015 at 12:06 | comment | added | Hachino | Riesz's projection theorem does not depend on the choice of a basis, so I don't see your point ? Take $Q_n$ to be the orthogonal projection on $V_n$ and your two conditions are satisfied (with $C = 1$ btw). | |
| Apr 1, 2015 at 11:58 | history | asked | Charpe | CC BY-SA 3.0 |