Metamath Proof Explorer


Theorem fourierdlem9

Description: H is a complex function. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses fourierdlem9.f ( 𝜑𝐹 : ℝ ⟶ ℝ )
fourierdlem9.x ( 𝜑𝑋 ∈ ℝ )
fourierdlem9.r ( 𝜑𝑌 ∈ ℝ )
fourierdlem9.w ( 𝜑𝑊 ∈ ℝ )
fourierdlem9.h 𝐻 = ( 𝑠 ∈ ( - π [,] π ) ↦ if ( 𝑠 = 0 , 0 , ( ( ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) − if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ) / 𝑠 ) ) )
Assertion fourierdlem9 ( 𝜑𝐻 : ( - π [,] π ) ⟶ ℝ )

Proof

Step Hyp Ref Expression
1 fourierdlem9.f ( 𝜑𝐹 : ℝ ⟶ ℝ )
2 fourierdlem9.x ( 𝜑𝑋 ∈ ℝ )
3 fourierdlem9.r ( 𝜑𝑌 ∈ ℝ )
4 fourierdlem9.w ( 𝜑𝑊 ∈ ℝ )
5 fourierdlem9.h 𝐻 = ( 𝑠 ∈ ( - π [,] π ) ↦ if ( 𝑠 = 0 , 0 , ( ( ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) − if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ) / 𝑠 ) ) )
6 0red ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ 𝑠 = 0 ) → 0 ∈ ℝ )
7 1 adantr ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → 𝐹 : ℝ ⟶ ℝ )
8 2 adantr ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → 𝑋 ∈ ℝ )
9 pire π ∈ ℝ
10 9 renegcli - π ∈ ℝ
11 iccssre ( ( - π ∈ ℝ ∧ π ∈ ℝ ) → ( - π [,] π ) ⊆ ℝ )
12 10 9 11 mp2an ( - π [,] π ) ⊆ ℝ
13 12 sseli ( 𝑠 ∈ ( - π [,] π ) → 𝑠 ∈ ℝ )
14 13 adantl ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → 𝑠 ∈ ℝ )
15 8 14 readdcld ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → ( 𝑋 + 𝑠 ) ∈ ℝ )
16 7 15 ffvelrnd ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) ∈ ℝ )
17 16 adantr ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) ∈ ℝ )
18 3 4 ifcld ( 𝜑 → if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ∈ ℝ )
19 18 ad2antrr ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ∈ ℝ )
20 17 19 resubcld ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → ( ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) − if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ) ∈ ℝ )
21 14 adantr ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → 𝑠 ∈ ℝ )
22 neqne ( ¬ 𝑠 = 0 → 𝑠 ≠ 0 )
23 22 adantl ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → 𝑠 ≠ 0 )
24 20 21 23 redivcld ( ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) ∧ ¬ 𝑠 = 0 ) → ( ( ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) − if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ) / 𝑠 ) ∈ ℝ )
25 6 24 ifclda ( ( 𝜑𝑠 ∈ ( - π [,] π ) ) → if ( 𝑠 = 0 , 0 , ( ( ( 𝐹 ‘ ( 𝑋 + 𝑠 ) ) − if ( 0 < 𝑠 , 𝑌 , 𝑊 ) ) / 𝑠 ) ) ∈ ℝ )
26 25 5 fmptd ( 𝜑𝐻 : ( - π [,] π ) ⟶ ℝ )