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I am considering the Banach space $A$ with $\sup$-norm, which is the uniform closure of functions on a segment that are continuous but a finite collection of jump points, where they have limits from the left and from the right. It looks like any finite complex measure on the segment naturally generates a functional on $A$: on the dense set of piecewise continuous functions one can take the usual integral (if a jump point has a point mass, we may split this mass into two equal parts for the left and right parts of the function). Is it true that the space of measures is dual to $A$?

My own comment: I see that the answer is negative: fix a point of the segment and take a number $c$, consider the functional that assigns to a function whose left and right limits at this point are $a_+$ and $a_-$ the number $c\cdot a_+ + (1-c)a_-$. On continuous functions this functional does not depend on $c$.

I would like to extend my question as follows: describe the space dual to $A$.

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I don't know if I should change the title or what; please feel free to edit. –  peno Aug 21 '11 at 20:41
    
As far as I understand, I can answer my question myself, but the system says that I can do this only in 5 hours. To construct a functional, take a (finite complex) measure which will define the functional on continuous functions; also take two pure point measures whose sum will be equal to the pure point part of our measure, and for piecewise continuous functions use them to deal with jumps, namely, integrate the left limits of the functions with respect to the first measure and the right ones with respect to the second. This uniquely defines the bounded functional on the dense set. –  peno Aug 21 '11 at 21:19

1 Answer 1

up vote 1 down vote accepted

Perhaps not quite right, but I leave it anyway

Consider $T$, the "two arrows space" (aka "top and bottom of the lexicographic square"). It has two points for each real in the interior of $[0,1]$ which can be called the "left" and "right" versions of that real. So $T$ is a compact Hausdorff non-metrizable topological space. Your normed space $A$ is naturally identified with (a subspace of?) the set of real-valued continuous functions on $T$, so that the dual is the measures on $T$.

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Thanks for the nice model of my space $A$. If I understand it correctly, continuous functions on $T$ correspond to functions on $A$ that have limits from the left (values on the bottom) and from the right (on the top). –  peno Aug 22 '11 at 4:48
    
For the two-arrows space you do not want to allow values at a point different from the two one-sided limits. So you may use CADLAG functions for example. CADLAG = "right-continuous with left limits", French initials. –  GEdgar Aug 22 '11 at 14:52
    
I meant the way to see the values of continuous functions on the compact. –  peno Aug 23 '11 at 4:19

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