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visits | member for | 2 years, 4 months |
seen | Jun 21 '13 at 22:27 | |
stats | profile views | 569 |
Dec 5 |
awarded | Necromancer |
Dec 3 |
awarded | Necromancer |
Aug 28 |
awarded | Yearling |
Aug 28 |
awarded | Yearling |
Aug 8 |
awarded | Necromancer |
Jun 25 |
awarded | Revival |
Jun 9 |
comment |
Defining definite integral using indefinite integral.
This maybe not pertinent, but there is a very general version if one is prepared to use the framework of distribution theory. Every distribution has a primitive and one then can use the concept of the limit of a distribution at a point or at infinity (which can be defined in an elementary fashion, i.e., without functional analysis) to define definite integrals---the fundamental theorem then holds. For example, the integral of the $\delta$-distribution is $1$ as one would expect, more surprisingly that of $\cos x$ is zero. Details are in an article by J. Sebastião e Silva (MR0216289). |
Jun 9 |
comment |
Defining definite integral using indefinite integral.
The Cantor function (more picturesquely, the devil's staircae) is a continuous, non-zero function whose derivative vanishes almost everywhere---see the corresponding Wikipedia article. |
Jun 3 |
answered | Uniformities generated by metrics. |
Jun 2 |
comment |
When is the support of a Radon measure separable?
The reason I suggested using a group was to avoid such questions, since in this situation you can use the existence theorem for Haar measure (which, by the way, has a very short and transparent proof in the case of a compact group, using the weak star compactness of the dual ball of $C(K)$). A particularly simple solution to the OP is then provided by a large cardinal product of the two-point group. |
Jun 1 |
comment |
When is the support of a Radon measure separable?
What about Haar measure on a compact, non-separable group? (By the way, don't you mean "smallest closed set with full measure"?). |
May 30 |
awarded | Fanatic |
May 26 |
comment |
Unambiguous “weak” vector valued $L^{+\infty}$ spaces?
By the way, with suitable tools, one can prove the required representation for the dual in a few lines. One can express $E$ as the closure of the union of an increasing sequence $(E_n)$ of finite dimensional subspaces. It follows easily that the required $L^1$ space is the inductive limit (in the category of Banach spaces with linear contractions as morphisms) of the $L^(I,E_n)$. The natural extension of $L^1$ duality theory to functions with values in finite dimensional spaces plus abstract nonsense on duality for the above category produces the stated identification of the dual. |
May 26 |
comment |
Unambiguous “weak” vector valued $L^{+\infty}$ spaces?
Even if the dual of the separable Banach space is not separable, there is a perfectly respectable complete locally convex structure there for which the dual of $L^1(I,E)$ is canonically identifiable with the space of (equivalence classes of) of bounded, measurable functions--- the bounded, weak-star topology. This not unimportant since probably the most significant example is that where $E$ is the space of bounded, linear operators on Hilbert space which is a non-separable dual of the nuclear operators. There this concept of measurability is ubiquitous in spectral theory. |
May 26 |
comment |
Hahn-Banach restricted to a pre-dual
In the second case you have the dual of a Frechet space and the topology of uniform convergence on compacta fits the bill---(consult the theorem of Banach-Dieudonne). I am assuming that the phrase "compact support" refers to the measures. |
May 26 |
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Hahn-Banach restricted to a pre-dual
As mentioned above, you can allways take the weak-star topology on the dual and use the Hahn-Banach theroem in this context. However, you seem to want a more intrinsic topology in concrete situations. Normally, one would like a complete LC structure. This is certainly possible for most concrete examples, in particular, for the two mentioned here. In fact, the dual of any Banach space has such a topology---the so-called bounded weak-star topology (this takes care of your first example). |
May 26 |
comment |
Embedding Theorem for topological spaces, and in general
The concept of universa space is possibly what you are looking for. $X$ is universal for a class of topological space if each of them embeds into it as a clossed subspace. This plays an important role, particularly in descriptive topology. As a starting point you could consult "Analytic sets" by C.A. Rogers (ed.) for the following important universal spaces---countable products of the closed unit interval (already mentioned), the integers (i.e., the irrationals), the two point space (the Cantor set) and the real line. |
May 20 |
comment |
Existence of dominating measure for weak*-compact set of measures
By "compatibility" above I of course meant "non-compatibility". It is perhaps worth mentioning that in the topological situation, the universal property works in the other direction. $S$ embeds into $M^t(S)$ in such a way that every continuous, bounded function on $S$ with values in a Banach space lifts in a unique fashion to a continuous linear mapping with the appropirate (which, again, is not the norm). |
May 20 |
comment |
Existence of dominating measure for weak*-compact set of measures
Compatibility in the first part means that the corresonding dual spaces are too large. Thus the Banach space duals of the function spaces consist in both cases of the finitely additive measures, not the countably additive or Radon ones. The unversal property is that every countably additive meaure on the $\sigma$-algebra with values in a Banach space (for which see Diestel and Uhl "Vector measures") lifts to a unique continuous linear mapping on $L^\infty$ with the topology mentioned in my answer. |
May 20 |
revised |
Existence of dominating measure for weak*-compact set of measures
corrected some irritating typos. |