# Why does it seem that $rca=rba$? [closed]

The following paradox has got me stumped. I'm hoping someone can point out the error.

Take a locally compact metric space $X$ and define the $C_b(X)$ and $C_0(X)$ as the spaces of continuous real-valued functions on $X$, bounded for $C_b$ and converging to zero at infinity for $C_0$, and equip the spaces with the supremum norm.

By Dunford & Schwartz Th. IV.6.2, the topological dual $C_b'$ is isometrically isomorphic to the space $rba(X)$ of regular bounded finitely additive Borel measures on $X$; by Rudin (Real and Complex Analysis, Th. 6.19) $C_0'$ is isometrically isomorphic to the space $rca(X)$ of regular countably additive Borel measures. In both cases the identification has the form $\langle \xi,f\rangle = \int_X f\, d\mu_\xi$.

Since $C_0\subset C_b$, and since $C_0$ and $C_b$ share the same norm, a simple calculation gives that $C_b'\subset C_0'$; since the two identification structures above are the same, this implies that $rba\subset rca$.

However, the definitions of $rba$ and $rca$ immediately imply that $rca\subset rba$.

What is going wrong here?

• The map $C'_b\to C'_0$ ("restriction") induced by the inclusion $C_0 \to C_b$ is not generally injective. – Sean Eberhard Jul 14 '14 at 10:47
• 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" – Yemon Choi Jul 14 '14 at 11:07
• @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 $c_0$. – Sean Eberhard Jul 14 '14 at 15:10
• 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. – Nate Eldredge Jul 14 '14 at 15:15
• This question appears to be off-topic because it is based on a natural but elementary error. – Yemon Choi Jul 14 '14 at 17:59