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Embedding L1 in the Banach space of complex Borel measures

  1. Jun 11, 2012 #1
    Hey, I know this is commonly a homework question, but it came up in my own studies; so this isn't a homework question for me. I hope it's alright that I put it here.

    I'm trying to show that if [itex] f dx = d\lambda[/itex] for some [itex] f \in L^1(\mathbb{R}^d) [/itex] and complex Borel measure [itex]\lambda[/itex] then [itex] |f| dx = d|\lambda| [/itex] (i.e., [itex] f \to fdx [/itex] is an isometry).

    What I've done so far is constructed a [itex] \psi \in L^1(\lambda) [/itex] so that [itex] \psi d\lambda = d|\lambda| [/itex] with [itex]|\psi| = 1 [/itex]. So we then have that [itex] f \psi dx = \psi d\lambda = d|\lambda| [/itex], so it looks like what I need to show is that [itex] f \psi = |f| [/itex]. I'm not really sure how to do that though.
  2. jcsd
  3. Jun 14, 2012 #2
    Oh dear, sorry, what I meant is that I have a [itex] \psi \in L^1(|\lambda|) [/itex] so that [itex] d\lambda = \psi d|\lambda|[/itex]. Sorry for the mistake.

    Anyway, what I've ended up doing so far is showing one of the two ineqaulities. Fix a partition [itex]\{X_j\}_{j=1}^{\infty} [/itex] of [itex] \mathbb{R}^d[/itex].

    Then [itex] \sum_{j=1}^{\infty} |\lambda(X_j)| = \sum_j |\int_{X_j} f(x) dx | \le \sum_j \int_{x_j} |f| dx = \int_{\mathbb{R}^d} |f| dx = ||f||_1 [/itex]

    with the second to last equality holding from the monotone convergence theorem. Supping on all such partitions gives [itex] |\lambda| \le ||f||_1 [/itex] directly from the definition. Still at kind of a loss how to show the reverse.
  4. Jun 20, 2012 #3
    For completeness, I thought I'd go ahead and update this with the solution.

    First I'll need the following fact:

    If [itex] \mu [/itex] is some positive measure, [itex] g \in L^1(\mu) [/itex], and for every measurable set [itex] E[/itex], [itex] 0 \le \int_E g(x) d\mu(x) [/itex], then [itex] g \ge 0 [/itex] almost everywhere (wrt [itex] \mu [/itex]).

    Proof: Consider the set [itex] F = \{ g(x) \notin [0, \infty) \} [/itex]. Write [itex] F = R \cup C[/itex] for [itex] R = \{ g(x) < 0 \} [/itex] and [itex] C = \{g(x) \in \mathbb{C} - \mathbb{R} \} [/itex]. F has positive measure if and only if at least one of R or C does as well.

    Suppose R has positive measure. Write [itex] R = \bigcup_j R_j [/itex] for [itex] R_j = \{ g(x) < -\frac{1}{j} \} [/itex]. Since R has positive measure, there's at least one j so that [itex]R_j [/itex] does as well. Thus [itex]\int_{R_j} g(x) d\mu \le \int_{R_j} -\frac{1}{j} d\mu = -\frac{1}{j} \mu(R_j) < 0 [/itex], which contradicts the hypothesis on g. So R must have 0 measure.

    Write g = u + iv. So on C, v is never 0. So [itex] 0 \le \int_C g(x) d\mu = \int_C u d\mu + i \int_C v d\mu [/itex]. This is only possible if [itex] \int_C v d\mu = 0 [/itex]. By writing [itex] C = P \cup N [/itex] where [itex] P = \{v > 0 \} [/itex] and [itex] N = \{ v < 0 \} [/itex] and applying the same argument as above on [itex] R[/itex], we see that both P and N must have 0 measure, so C does as well.

    Together this shows that F must have 0 measure. //

    Onto the actual statement, we want to show that:

    If [itex] \mu [/itex] is a positive measure, [itex] f \in L^1(\mu) [/itex] and [itex] \lambda [/itex] is defined by [itex] d\lambda = f d\mu [/itex], then [itex] ||\lambda|| = |f| d\mu [/itex]

    Proof: Construct a function [itex] \psi \in L^1(|\lambda|) [/itex] with [itex] |\psi| = 1[/itex] and [itex] \psi d\|lambda| = d\lambda [/itex]. Now [itex] \psi \bar{\psi} = |\psi| = 1 [/itex], so

    [itex] d|\lambda| = \bar{\psi} \psi d|\lambda| = \bar{\psi} d\lambda = \bar{\psi} f d\mu [/itex].

    Thus given any measurable set E, we have that [itex] 0 \le |\lambda|(E) = \int_E \bar{\psi} f d\mu [/itex].

    Appealing to the fact above, this implies that [itex] \bar{\psi} f \ge 0 [/itex] almost everywhere (so modifying on a set of measure 0 and changing nothing in the above argument, we may assume it's nonnegative everywhere).

    Thus [itex] \bar{\psi} f = |\bar{\psi} f| = |f| [/itex], since [itex] |\bar{\psi} | = 1 [/itex]. So, [itex] d|\lambda| = \bar{\psi} f d\mu = |f| d\mu [/itex], which gives the result.
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