You put your right foot in, you take your right foot out
Any complex measure decomposes into an absolutely continuous
measure and a singular measure , with respect to some positive
measure . This is the Lebesgue decomposition,
RE: You put your right foot in, you take your right foot out
S. G. Kal'nei derived in [5], [6] a quite sharp necessary condition for the multiplier norm of a finite sequence in the setting of Fourier-Jacobi series on L 1 with "natural weight" (which ensures a nice convolution structure). In this paper, Kalnei's problem is considered in the setting of Laguerre series on weighted L 1 -spaces; the admitted scale of weights contains in particular the appropriate "natural weights" occurring in transplantation and convolution. [8D]
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The Tazman aka Martin Price
Proud father of a Devil Dog
RE: You put your right foot in, you take your right foot out
Quote:
ORIGINAL: Tazman
S. G. Kal'nei derived in [5], [6] a quite sharp necessary condition for the multiplier norm of a finite sequence in the setting of Fourier-Jacobi series on L 1 with "natural weight" (which ensures a nice convolution structure). In this paper, Kalnei's problem is considered in the setting of Laguerre series on weighted L 1 -spaces; the admitted scale of weights contains in particular the appropriate "natural weights" occurring in transplantation and convolution. [8D]
someone better check on Taz's background....he could be a foreign spy, cuz he sure aint talkin' English !!!!
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It' s a dog eat dog world, and I' m wearing Milkbone underwear
RE: You put your right foot in, you take your right foot out
The Lebesgue integral is strictly more general than the proper Riemann integral -- i.e., it can integrate a wider class of functions. However, in comparing the improper Riemann integral with the Lebesgue integral, we find that neither is strictly more general than the other.
Does that help you any priz?
__________________
The Tazman aka Martin Price
Proud father of a Devil Dog
RE: You put your right foot in, you take your right foot out
Quote:
The Lebesgue integral is strictly more general than the proper Riemann integral -- i.e., it can integrate a wider class of functions. However, in comparing the improper Riemann integral with the Lebesgue integral, we find that neither is strictly more general than the other.
I remember hearing some of that stuff in college, but I don't remember a single, specific thing now!!
Maybe we should request an advanced mathematics area here at HNI. It might be fun to brush up on some of this stuff!!
RE: You put your right foot in, you take your right foot out
etothepii do you recall this?
Quote:
Construction of the Lebesgue integral
Let μ be a (non-negative) measure on a sigma-algebra X over a set E. (In real analysis, E will typically be Euclidean n-space Rn or some Lebesgue measurable subset of it, X will be the sigma-algebra of all Lebesgue measurable subsets of E, and μ will be the Lebesgue measure. In probability and statistics, μ will be a probability measure on a probability space E.) We build up an integral for real-valued functions defined on E as follows.
Fix a set S in X (S ∈ X, S ⊂ E) and let f be the function on E whose value is 0 outside of S and 1 inside of S (i.e., f(x) = 1 if x is in S, otherwise f(x) = 0.) This is called the indicator function or characteristic function of S and is denoted 1S.
To assign a value to ∫1S consistent with the given measure μ, the only reasonable choice is to set: \int 1_S = \mu (S) We extend by linearity to the linear span of indicator functions: \int \sum a_k 1_{S_k} = \sum a_k \mu( S_k) where the sum is finite and the coefficients ak are real numbers. Such a finite linear combination of indicator functions is called a simple function. Note that a simple function can be written in many ways as a linear combination of characteristic functions, but the integral will always be the same.
Now the difficulties begin as we attempt to take limits so that we can integrate more general functions. It turns out that the following process works and is most fruitful.
Let f be a non-negative function supported on the set E (we allow it to attain the value +∞, in other words, f takes values in the extended real number line.) We define ∫f to be the supremum of ∫s where s varies over all simple functions which are under f (that is, s(x) ≤ f(x) for all x.) This is analogous to the lower sums of Riemann. However, we will not build an upper sum, and this fact is important in getting a more general class of integrable functions. One can be more explicit and mention the measure and domain of integration: \int_E f\,d\mu := \sup\left\{\,\int_E s\,d\mu : s\le f,\ s\ \mbox{simple}\,\right\} There is the question of whether this definition makes sense (do simple function or indicating function keep the same integral?) There is also the question of whether this corresponds in any way to a Riemann notion of integration. It is not so hard to prove that the answer to both questions is yes.
We have defined ∫f for any non-negative function on E; however for some functions ∫f will be infinite. Furthermore, desirable additive and limit properties of the integral are not satisfied, unless we require that all our functions are measurable, meaning that the pre-image of any interval is in X. We will make this assumption from now on.
To handle signed functions, we need a few more definitions. If f is a function of the measurable set E to the reals (including
∞), then we can write f = g − h where g(x) = (f(x) if f(x)>0, 0 otherwise) and h(x) = (−f(x) if f(x) < 0, 0 otherwise). Note that both g and h are non-negative functions. Also note that |f| = g + h. If ∫|f| is finite, then f is called Lebesgue integrable. In this case, both ∫g and ∫h are finite, and it makes sense to define ∫f by ∫g − ∫h. It turns out that this definition is the correct one. Complex valued functions can be similarly integrated, by considering the real part and the imaginary part separately.
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The Tazman aka Martin Price
Proud father of a Devil Dog
You put your right foot in, you take your right foot out
