Timeline for Why does it seem that $rca=rba$? [closed]
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| Jul 15, 2014 at 22:07 | history | closed |
Yemon Choi Stefan Kohl♦ Ryan Budney S. Carnahan♦ |
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| Jul 14, 2014 at 18:16 | review | Close votes | |||
| Jul 15, 2014 at 22:07 | |||||
| Jul 14, 2014 at 17:59 | comment | added | Yemon Choi | This question appears to be off-topic because it is based on a natural but elementary error. | |
| Jul 14, 2014 at 15:30 | comment | added | Yemon Choi | Just a quick addition to my previous comment: the magic words or jargon in this context are: "Yosida-Hewitt decomposition" | |
| Jul 14, 2014 at 15:20 | comment | added | Nate Eldredge | @SeanEberhard: Good example. Moreover, let's call your functional $f$. If $i$ is the inclusion $i : c_0 \to \ell^\infty$, let $i^* : (\ell^\infty)^* \to (c_0)^* = \ell^1$, and you can check that $i^* f = 0$. So we see explicitly that $i^*$ is not injective. | |
| Jul 14, 2014 at 15:15 | comment | added | Nate Eldredge | As an elementary mnemonic, if you have a linear map $T : X \to Y$ and $T^*$ is its adjoint, you should expect the injectivity of $T^*$ to be related to the surjectivity of $T$, and vice versa. (Think about matrices with some zero rows or columns.) It's a little trickier in infinite dimenions (e.g. in some cases, instead of "surjective" you want "dense range") but it helps in this case: since the inclusion from $C_0$ to $C_b$ is not surjective (nor does it even have dense range) you should not expect its adjoint to be injective. | |
| Jul 14, 2014 at 15:10 | comment | added | Sean Eberhard | @MarkPeletier Exactly. For example consider $X=\mathbf{N}$. Then $C_0 = c_0$ and $C_b = \ell^\infty$. Between these two spaces is the space $c$ of all sequences $(x_n)$ which converge to some limit. The map $(x_n)\mapsto \lim x_n$ is a bounded linear functional on $c$, and so by the Hahn-Banach theorem it extends to some nontrivial bounded linear functional on $\ell^\infty$ which vanishes on $c_0$. | |
| Jul 14, 2014 at 14:54 | comment | added | Mark Peletier | @YemonChoi, thinking about the injection as a quotient map makes things much clearer. Thanks! | |
| Jul 14, 2014 at 14:54 | comment | added | Mark Peletier | @SeanEberhard, that is very useful! As I understand it, $C_0$ is not dense in $C_b$, and therefore knowing $\xi\in C_b'$ on $C_0$ is not enough to determine $\xi$. | |
| Jul 14, 2014 at 11:07 | comment | added | Yemon Choi | As Sean Eberhard notes, what you get is a _quotient_map from rba onto rca. In fact rca is a direct summand of rba, and your quotient map is "throwing away the singular part" | |
| Jul 14, 2014 at 10:47 | comment | added | Sean Eberhard | The map $C'_b\to C'_0$ ("restriction") induced by the inclusion $C_0 \to C_b$ is not generally injective. | |
| Jul 14, 2014 at 10:18 | history | asked | Mark Peletier | CC BY-SA 3.0 |