I want to calculate the dual space of $\mathcal{C}_0[a,b]$, that is the space of continuous functions on $[a,b]$ vanishing at $a$. I know that the dual space of $\mathcal{C}[a,b]$ (continuous functions) is the space of differences of Lebesgue-Stieltjes measure associated to increasing functions, left-continuous, and vanishing at a fixed point in $[a,b]$. Anyone can help me?
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$\begingroup$ This is simple, or not? What you define $C_0[a,b]$ is usually referred to as $C_0(a,b]$ in the $C^*$-algebra literature. For $X$ loc.cpct.Hausdorff, the dual of $C_0(X)$ is well-known and explicit. $\endgroup$– Marc PalmCommented Mar 21, 2013 at 10:35
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3$\begingroup$ Dear filippo, Evaluation at $a$ gives a short exact sequence $$0 \to C_0[a,b] \to C[a,b] \to \mathbb C \to 0,$$ and so taking duals gives the short exact (by Hahn--Banach) sequence $$0 \to \mathbb C \to C[a,b]' \to C_0[a,b] ' \to 0.$$ This gives a fairly functorial point of view on njguliyev's answer below, and gives a general strategy for answering this kind of question, where you modify a function space you already understand by some sort of vanishing condition. Regards, $\endgroup$– EmertonCommented Mar 21, 2013 at 12:09
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$\begingroup$ Cross-posted and answered at math.stackexchange.com/questions/336737/… $\endgroup$– JRNCommented Mar 27, 2013 at 23:00
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2 Answers
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Each (bounded linear) functional on $C[a,b]$ is also a functional on $C_0[a,b]$. Each function $f \in C[a,b]$ can be written as $f(x) = f_0(x) + f(a) \cdot 1$, where $f_0 \in C_0[a,b]$. On the other hand, each functional on $C_0[a,b]$ can be extended to a functional on $C[a,b]$ (Hahn-Banach). Therefore the answer is $(C[a,b])^{\ast}/\mathbb{C}$. In terms of those functions of bounded variation, this means that we have to require those functions vanish at one more point.
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1$\begingroup$ Extending a functional by one dimension only does not require Hahn-Banach. But that language still may be a useful one for visualizing it. $\endgroup$ Commented Mar 21, 2013 at 12:24
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$\begingroup$ Of course. I didn't notice that... $\endgroup$– user26107Commented Mar 21, 2013 at 12:52
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$C_0[a,b]^* = C[a,b]^*/C_0[a,b]^{ann}$ where $C_0[a,b]^{ann}$ is the annihilator of $C_0[a,b]$ in $C_0[a,b]^*$; the annihilator is 1-dimensional with basis $\delta_a$, the atomic measure at $a$, or the evaluation at $a$. In your setup this corresponds to the Stieltjes integral with respect to the function which is 1 at $a$ and 0 elsewhere. If your fixed point in $[a,b]$ is $a$ you can take instead the function which is 0 at $a$ and is 1 elsewhere which gives $-\delta_a$ as Stieltjes integral.
answered Mar 21, 2013 at 10:32