Metamath Proof Explorer


Theorem cnlnadjlem9

Description: Lemma for cnlnadji . F provides an example showing the existence of a continuous linear adjoint. (Contributed by NM, 18-Feb-2006) (New usage is discouraged.)

Ref Expression
Hypotheses cnlnadjlem.1 𝑇 ∈ LinOp
cnlnadjlem.2 𝑇 ∈ ContOp
cnlnadjlem.3 𝐺 = ( 𝑔 ∈ ℋ ↦ ( ( 𝑇𝑔 ) ·ih 𝑦 ) )
cnlnadjlem.4 𝐵 = ( 𝑤 ∈ ℋ ∀ 𝑣 ∈ ℋ ( ( 𝑇𝑣 ) ·ih 𝑦 ) = ( 𝑣 ·ih 𝑤 ) )
cnlnadjlem.5 𝐹 = ( 𝑦 ∈ ℋ ↦ 𝐵 )
Assertion cnlnadjlem9 𝑡 ∈ ( LinOp ∩ ContOp ) ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝑡𝑧 ) )

Proof

Step Hyp Ref Expression
1 cnlnadjlem.1 𝑇 ∈ LinOp
2 cnlnadjlem.2 𝑇 ∈ ContOp
3 cnlnadjlem.3 𝐺 = ( 𝑔 ∈ ℋ ↦ ( ( 𝑇𝑔 ) ·ih 𝑦 ) )
4 cnlnadjlem.4 𝐵 = ( 𝑤 ∈ ℋ ∀ 𝑣 ∈ ℋ ( ( 𝑇𝑣 ) ·ih 𝑦 ) = ( 𝑣 ·ih 𝑤 ) )
5 cnlnadjlem.5 𝐹 = ( 𝑦 ∈ ℋ ↦ 𝐵 )
6 1 2 3 4 5 cnlnadjlem6 𝐹 ∈ LinOp
7 1 2 3 4 5 cnlnadjlem8 𝐹 ∈ ContOp
8 elin ( 𝐹 ∈ ( LinOp ∩ ContOp ) ↔ ( 𝐹 ∈ LinOp ∧ 𝐹 ∈ ContOp ) )
9 6 7 8 mpbir2an 𝐹 ∈ ( LinOp ∩ ContOp )
10 1 2 3 4 5 cnlnadjlem5 ( ( 𝑧 ∈ ℋ ∧ 𝑥 ∈ ℋ ) → ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) )
11 10 ancoms ( ( 𝑥 ∈ ℋ ∧ 𝑧 ∈ ℋ ) → ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) )
12 11 rgen2 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) )
13 fveq1 ( 𝑡 = 𝐹 → ( 𝑡𝑧 ) = ( 𝐹𝑧 ) )
14 13 oveq2d ( 𝑡 = 𝐹 → ( 𝑥 ·ih ( 𝑡𝑧 ) ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) )
15 14 eqeq2d ( 𝑡 = 𝐹 → ( ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝑡𝑧 ) ) ↔ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) ) )
16 15 2ralbidv ( 𝑡 = 𝐹 → ( ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝑡𝑧 ) ) ↔ ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) ) )
17 16 rspcev ( ( 𝐹 ∈ ( LinOp ∩ ContOp ) ∧ ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝐹𝑧 ) ) ) → ∃ 𝑡 ∈ ( LinOp ∩ ContOp ) ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝑡𝑧 ) ) )
18 9 12 17 mp2an 𝑡 ∈ ( LinOp ∩ ContOp ) ∀ 𝑥 ∈ ℋ ∀ 𝑧 ∈ ℋ ( ( 𝑇𝑥 ) ·ih 𝑧 ) = ( 𝑥 ·ih ( 𝑡𝑧 ) )