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Lecture 32 - Apr 11

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Continuation of Lecture 31

To finish the proof that for , use the inequalities obtained there:

  1. From the left-interval family,
  1. From the right-interval family,
  1. Comparing both sums (monotonicity + disjointness argument) gives

Letting ,

Since , we must have

Hence , and the monotone function is differentiable a.e.

Bounded Variation

Let and let

be a partition of . For , define the variation of with respect to by

Define total variation:

If , we say is of bounded variation (BV) on .

Examples

  1. If is BV on , then is bounded.
  2. Denote by the set of all BV functions on ; it is a vector space.

Lemma

Suppose is BV. Then:

  1. For any ,
  1. The function

is increasing.

  1. The function

is increasing.

Proof

For (1): if is a partition of containing , then

Taking supremum over partitions gives

For reverse inequality, choose partitions that are -close to , merge them, and let .

For (2): if and is any partition of , then is a partition of . Hence

Taking supremum over gives

so is increasing.

Also from the same inequality,

thus

which is equivalent to

So (3) follows.

Remark (Jordan Idea)

The monotonicity of

is the key ingredient for writing a BV function as a difference of two increasing functions.

Examples

is continuous but not of bounded variation on .

is continuous and of bounded variation on .

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