Lecture 23 - Mar 11

Continuation: Regularity and Start of Preliminaries

This lecture continues the theorem from Theorem.

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Theorem (Recap)

Let be a metric space and

be a measure. Then:

1. For every ,

2. The same formulas hold for every .

3. If is -compact, then

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Proof of (3) (continued)

Assume

where each is compact, and (without loss) .

Fix and . By (2), choose closed such that

Since , by continuity from below there is such that

Hence

Now is compact and contained in , so taking supremum over compact subsets of gives the claim.

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Remarks

1. is not -compact (usual metric).
2. Exercise: a connected locally compact space is -compact.

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Complex-Valued Measurable and Integrable Functions

Let be a measure space and .

Definition

1. is measurable iff and are measurable real-valued functions.
2. is integrable iff and are integrable.

In that case, define

Basic properties

If are integrable and , then

Also, pointwise,

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Proposition

is integrable iff is integrable.

Moreover,

Proof sketch of the inequality

If , done. Otherwise define

Then

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Convexity Reminder

A real-valued function on an interval is convex if for all and ,

For concave functions, the inequality reverses.

Examples:

1. () is convex on .
2. is concave on .

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Lemma (Preliminaries for Theory)

Let , and define the conjugate exponent by

Then:

1. For ,

2. (Young’s inequality) For ,

Idea of proofs

1. Use convexity of at the midpoint:

2. Use concavity of with

Then

and exponentiating gives Young’s inequality.