Exercise 15.2.2

Supply the details needed to complete the proof of Proposition 15.2.1.

Answers

Proof. The proof of Proposition 15.2.1 shows that

φ ′ ( s ) = 1 − φ 4 ( s ) , 0 ≤ s ≤ ϖ 2 .

By Exercise 3, parts (a) and (b), φ ′ is even and has period 2 ϖ , and by part (c),

φ ′ ( ϖ − s ) = − φ ′ ( s ) , s ∈ ℝ .

Therefore, if − ϖ 2 ≤ s ≤ 0 , then

φ ′ ( s ) = − φ ′ ( − s ) = − 1 − φ 4 ( − s ) = − 1 − φ 4 ( s ) .

Now, if ϖ 2 ≤ s ≤ ϖ , then 0 ≤ ϖ − s ≤ ϖ 2 , thus

φ ′ ( s ) = − φ ′ ( ϖ − s ) = − 1 − φ 4 ( ϖ − s ) = − 1 − φ 4 ( s ) .

If − ϖ ≤ s ≤ − ϖ 2 , then ϖ 2 ≤ − s ≤ ϖ . Using the above equality, we obtain

φ ′ ( s ) = − φ ( − s ) = − 1 − φ 4 ( − s ) = − 1 − φ 4 ( s ) .

We have proved

φ ′ 2 ( s ) = 1 − φ 4 ( s ) , − ϖ ≤ s ≤ ϖ .

Now if s is any real number, there is some n ∈ ℤ and s ′ ∈ [ − ϖ , ϖ [ such that s = 2 nϖ + s ′ . Since 2 ϖ is a period of φ and φ ′ ,

φ ′ 2 ( s ) = φ ′ 2 ( s ′ ) = 1 − φ 4 ( s ′ ) = 1 − φ 4 ( s ) .

This complete the proof of Proposition 15.2.1. □

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2022-07-19 00:00
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