Metamath Proof Explorer


Theorem nmfn0

Description: The norm of the identically zero functional is zero. (Contributed by NM, 25-Apr-2006) (New usage is discouraged.)

Ref Expression
Assertion nmfn0 ( normfn ‘ ( ℋ × { 0 } ) ) = 0

Proof

Step Hyp Ref Expression
1 0lnfn ( ℋ × { 0 } ) ∈ LinFn
2 lnfnf ( ( ℋ × { 0 } ) ∈ LinFn → ( ℋ × { 0 } ) : ℋ ⟶ ℂ )
3 nmfnval ( ( ℋ × { 0 } ) : ℋ ⟶ ℂ → ( normfn ‘ ( ℋ × { 0 } ) ) = sup ( { 𝑥 ∣ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) } , ℝ* , < ) )
4 1 2 3 mp2b ( normfn ‘ ( ℋ × { 0 } ) ) = sup ( { 𝑥 ∣ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) } , ℝ* , < )
5 c0ex 0 ∈ V
6 5 fvconst2 ( 𝑦 ∈ ℋ → ( ( ℋ × { 0 } ) ‘ 𝑦 ) = 0 )
7 6 fveq2d ( 𝑦 ∈ ℋ → ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) = ( abs ‘ 0 ) )
8 abs0 ( abs ‘ 0 ) = 0
9 7 8 eqtrdi ( 𝑦 ∈ ℋ → ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) = 0 )
10 9 eqeq2d ( 𝑦 ∈ ℋ → ( 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ↔ 𝑥 = 0 ) )
11 10 anbi2d ( 𝑦 ∈ ℋ → ( ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) ↔ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = 0 ) ) )
12 11 rexbiia ( ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) ↔ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = 0 ) )
13 ax-hv0cl 0 ∈ ℋ
14 0le1 0 ≤ 1
15 fveq2 ( 𝑦 = 0 → ( norm𝑦 ) = ( norm ‘ 0 ) )
16 norm0 ( norm ‘ 0 ) = 0
17 15 16 eqtrdi ( 𝑦 = 0 → ( norm𝑦 ) = 0 )
18 17 breq1d ( 𝑦 = 0 → ( ( norm𝑦 ) ≤ 1 ↔ 0 ≤ 1 ) )
19 18 rspcev ( ( 0 ∈ ℋ ∧ 0 ≤ 1 ) → ∃ 𝑦 ∈ ℋ ( norm𝑦 ) ≤ 1 )
20 13 14 19 mp2an 𝑦 ∈ ℋ ( norm𝑦 ) ≤ 1
21 r19.41v ( ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = 0 ) ↔ ( ∃ 𝑦 ∈ ℋ ( norm𝑦 ) ≤ 1 ∧ 𝑥 = 0 ) )
22 20 21 mpbiran ( ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = 0 ) ↔ 𝑥 = 0 )
23 12 22 bitri ( ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) ↔ 𝑥 = 0 )
24 23 abbii { 𝑥 ∣ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) } = { 𝑥𝑥 = 0 }
25 df-sn { 0 } = { 𝑥𝑥 = 0 }
26 24 25 eqtr4i { 𝑥 ∣ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) } = { 0 }
27 26 supeq1i sup ( { 𝑥 ∣ ∃ 𝑦 ∈ ℋ ( ( norm𝑦 ) ≤ 1 ∧ 𝑥 = ( abs ‘ ( ( ℋ × { 0 } ) ‘ 𝑦 ) ) ) } , ℝ* , < ) = sup ( { 0 } , ℝ* , < )
28 xrltso < Or ℝ*
29 0xr 0 ∈ ℝ*
30 supsn ( ( < Or ℝ* ∧ 0 ∈ ℝ* ) → sup ( { 0 } , ℝ* , < ) = 0 )
31 28 29 30 mp2an sup ( { 0 } , ℝ* , < ) = 0
32 4 27 31 3eqtri ( normfn ‘ ( ℋ × { 0 } ) ) = 0