|analytic integration||cohomological integration|
|measurable space||Poincaré duality|
|measure||orientation in generalized cohomology|
|volume form||(virtual) fundamental class|
|Riemann/Lebesgue integration of differential forms||push-forward in generalized cohomology/in differential cohomology|
The term ‘Lebesgue space’ can stand for two distinct notions: one is the general notion of measure space (compare the Springer Encyclopaedia of Mathematics) and another is the notion of space (or space). Here we discuss the latter.
Sometimes ‘’ is used as a synonym for ; sometimes it is used to mean .
If is a real number and is a measure space, one considers the space of (equivalence classes of) measurable (complex- or real-valued) functions whose (absolute values of) th powers are integrable; i.e. whose norm
For fixed , the norm is continuous in . Accordingly, for , one may take the limit of as . However, this turns out to be the same as the essential supremum norm . Therefore, makes sense as long as is a measurable space equipped with a family of null sets (or full sets); the measure is otherwise irrelevant.
For , we can define the space using the same formula; however, it is no longer a Banach space (but still an F-space?). For , we can again take the limit, or equivalently use the essential infimum; every measurable function on belongs to (which is no longer even an -space).
We offer here a proof that defines a norm in the case ; the cases and follow by continuity and are easy to check from first principles. The most usual textbook proofs involve a clever application of Hölder's inequality?; the following proof is more straightforwardly geometric. All functions may be assumed to be real- or complex-valued.
Suppose , and suppose is a measure space with measure . Then the function defined by
defines a norm.
One must verify three things:
Separation axiom: implies .
Scaling axiom: .
Triangle inequality: .
The first two properties are obvious, so it remains to prove the last, which is also called Minkowski’s inequality.
Our proof of Minkowski’s inequality is broken down into a series of simple lemmas. The plan is to boil it down to two things: the scaling axiom, and convexity of the function (as a function from real or complex numbers to nonnegative real numbers).
First, some generalities. Let be a (real or complex) vector space equipped with a function that satisfies the scaling axiom: for all scalars , and the separation axiom: implies . As usual, we define the unit ball? in to be
Given that the scaling and separation axioms hold, the following conditions are equivalent: 1. The triangle inequality is satisfied. 2. The unit ball is convex. 3. If , then for all .
Condition 1. implies condition 2. easily: if and are in the unit ball and , we have
Now 2. implies 3. trivially, so it remains to prove that 3. implies 1. Suppose . Let and be the associated unit vectors. Then
where . If condition 3. holds, then
but by the scaling axiom, this is the same as saying
which is the triangle inequality.
Consider now with its -norm . By Lemma 1, this inequality is equivalent to
This allows us to remove the cumbersome exponent in the definition of the -norm.
The next two lemmas may be proven by elementary calculus; we omit the proofs. (But you can also see the full details.)
Let be two complex numbers, and define
for real . Then is nonnegative.
Define by . Then is convex, i.e., for all ,
for all .
Let and be unit vectors in . By condition 4, it suffices to show that for all . But
by Lemma 3. Using , we are done.
W. Rudin, Functional analysis, McGraw Hill 1991.
L. C. Evans, Partial differential equations, Amer. Math. Soc. 1998.
Wikipedia (English): Lp space