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Let $g: \mathbb{R} \to \mathbb{R}$ be a positive function uniformly bounded away from $0$. Let $C(\mathbb{R})$ be the space of continuous functions that with norm $| f | := \sup_{x \in \mathbb{R} } |f(x)|/ g(x)$, so it contains all bounded and continuous functions, and also functions that don't grow too fast. What is the norm dual of this space?

Victor
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Let $C_b(\mathbb R)$ be the space of all continuous bounded functions on $\mathbb R$. We have the isometric isomorphism $T:C(\mathbb R)\to C_b(\mathbb R)$ defined by $T(f)=f/g$. Therefore, the adjoint $T^*$ is an isometric isomorphism of $C_b(\mathbb R)^*$ onto $C(\mathbb R)^*$. In other words, the problem reduces to $g\equiv 1$.

Unfortunately, $C_b(\mathbb R)^*$ is not a nice space, but apparently there is a form of the Riesz representation for it (finitely additive measures).

  • I believe something more has to be done. Do you have a reference to statement $C(\mathbb{R})^*=" the space of finite measures. – AD - Stop Putin - Mar 05 '13 at 05:02
  • @AD. You are right, of course. I hope the new version makes more sense. –  Mar 05 '13 at 05:25
  • @5pm@AD Thanks to you both. Reducing the problem to $C_b(\mathbb{R})$ is useful, because the dual of this space is the space of finitely additive measures. Dunford and Schwartz describe this version of the Riesz representation theorem. - Victor – Victor Mar 05 '13 at 13:06
  • @5pm. Just to clarify, the Riesz representation theorem says that the dual $C_b(\mathbb{R})$ is the space of all regular, finitely additive signed measures on $\mathbb{R}$. (Regularity is very important.) With that being said, I think we can now conclude that the dual of $C(\mathbb{R})$ is the space of all regular, finitely additive measures $\mu$ on $\mathbb{R}$ that satisfy $|\mu| g < \infty$, ie, all measures such that the expectation of $g$ with respect to such a measure is finite. – Victor Mar 05 '13 at 17:19
  • @Victor Thanks for the clarification; I'm quite ignorant in that area (as the first version on my answer showed). If you want, you can edit my answer to make it more precise. –  Mar 05 '13 at 17:26
  • @5pm. I just want to make sure I understand the bit about isometries (before editing to get a final answer). What is the isometry between $C_b(\mathbb{R})^$ and $C(\mathbb{R})^$? I can see the isometry between the primal spaces, but not between the dual spaces. – Victor Mar 05 '13 at 21:40
  • @Victor If $T:X\to Y$ is an isometric isomorphism, then the adjoint of $T$, defined by $T^(\phi)=\phi\circ T$, is also an isometric isomorphism. If $\mu$ is a regular finitely additive signed measure on $\mathbb R$, then $T^(\mu)$ acts on $C(\mathbb R)$ by $\int T(f),d\mu$, which is simply $\int (f/g),d\mu$. You can summarize it as $T^*\mu=(g^{-1})\mu$. –  Mar 05 '13 at 21:45