Metamath Proof Explorer


Theorem nmo0

Description: The operator norm of the zero operator. (Contributed by Mario Carneiro, 20-Oct-2015)

Ref Expression
Hypotheses nmo0.1 N = S normOp T
nmo0.2 V = Base S
nmo0.3 0 ˙ = 0 T
Assertion nmo0 S NrmGrp T NrmGrp N V × 0 ˙ = 0

Proof

Step Hyp Ref Expression
1 nmo0.1 N = S normOp T
2 nmo0.2 V = Base S
3 nmo0.3 0 ˙ = 0 T
4 eqid norm S = norm S
5 eqid norm T = norm T
6 eqid 0 S = 0 S
7 simpl S NrmGrp T NrmGrp S NrmGrp
8 simpr S NrmGrp T NrmGrp T NrmGrp
9 ngpgrp S NrmGrp S Grp
10 ngpgrp T NrmGrp T Grp
11 3 2 0ghm S Grp T Grp V × 0 ˙ S GrpHom T
12 9 10 11 syl2an S NrmGrp T NrmGrp V × 0 ˙ S GrpHom T
13 0red S NrmGrp T NrmGrp 0
14 0le0 0 0
15 14 a1i S NrmGrp T NrmGrp 0 0
16 3 fvexi 0 ˙ V
17 16 fvconst2 x V V × 0 ˙ x = 0 ˙
18 17 ad2antrl S NrmGrp T NrmGrp x V x 0 S V × 0 ˙ x = 0 ˙
19 18 fveq2d S NrmGrp T NrmGrp x V x 0 S norm T V × 0 ˙ x = norm T 0 ˙
20 5 3 nm0 T NrmGrp norm T 0 ˙ = 0
21 20 ad2antlr S NrmGrp T NrmGrp x V x 0 S norm T 0 ˙ = 0
22 19 21 eqtrd S NrmGrp T NrmGrp x V x 0 S norm T V × 0 ˙ x = 0
23 2 4 nmcl S NrmGrp x V norm S x
24 23 ad2ant2r S NrmGrp T NrmGrp x V x 0 S norm S x
25 24 recnd S NrmGrp T NrmGrp x V x 0 S norm S x
26 25 mul02d S NrmGrp T NrmGrp x V x 0 S 0 norm S x = 0
27 14 26 breqtrrid S NrmGrp T NrmGrp x V x 0 S 0 0 norm S x
28 22 27 eqbrtrd S NrmGrp T NrmGrp x V x 0 S norm T V × 0 ˙ x 0 norm S x
29 1 2 4 5 6 7 8 12 13 15 28 nmolb2d S NrmGrp T NrmGrp N V × 0 ˙ 0
30 1 nmoge0 S NrmGrp T NrmGrp V × 0 ˙ S GrpHom T 0 N V × 0 ˙
31 12 30 mpd3an3 S NrmGrp T NrmGrp 0 N V × 0 ˙
32 1 nmocl S NrmGrp T NrmGrp V × 0 ˙ S GrpHom T N V × 0 ˙ *
33 12 32 mpd3an3 S NrmGrp T NrmGrp N V × 0 ˙ *
34 0xr 0 *
35 xrletri3 N V × 0 ˙ * 0 * N V × 0 ˙ = 0 N V × 0 ˙ 0 0 N V × 0 ˙
36 33 34 35 sylancl S NrmGrp T NrmGrp N V × 0 ˙ = 0 N V × 0 ˙ 0 0 N V × 0 ˙
37 29 31 36 mpbir2and S NrmGrp T NrmGrp N V × 0 ˙ = 0