Metamath Proof Explorer


Theorem lclkrs

Description: The set of functionals having closed kernels and majorizing the orthocomplement of a given subspace R is a subspace of the dual space. TODO: This proof repeats large parts of the lclkr proof. Do we achieve overall shortening by breaking them out as subtheorems? Or make lclkr a special case of this? (Contributed by NM, 29-Jan-2015)

Ref Expression
Hypotheses lclkrs.h H = LHyp K
lclkrs.o ˙ = ocH K W
lclkrs.u U = DVecH K W
lclkrs.s S = LSubSp U
lclkrs.f F = LFnl U
lclkrs.l L = LKer U
lclkrs.d D = LDual U
lclkrs.t T = LSubSp D
lclkrs.c C = f F | ˙ ˙ L f = L f ˙ L f R
lclkrs.k φ K HL W H
lclkrs.r φ R S
Assertion lclkrs φ C T

Proof

Step Hyp Ref Expression
1 lclkrs.h H = LHyp K
2 lclkrs.o ˙ = ocH K W
3 lclkrs.u U = DVecH K W
4 lclkrs.s S = LSubSp U
5 lclkrs.f F = LFnl U
6 lclkrs.l L = LKer U
7 lclkrs.d D = LDual U
8 lclkrs.t T = LSubSp D
9 lclkrs.c C = f F | ˙ ˙ L f = L f ˙ L f R
10 lclkrs.k φ K HL W H
11 lclkrs.r φ R S
12 ssrab2 f F | ˙ ˙ L f = L f ˙ L f R F
13 12 a1i φ f F | ˙ ˙ L f = L f ˙ L f R F
14 9 a1i φ C = f F | ˙ ˙ L f = L f ˙ L f R
15 eqid Base D = Base D
16 1 3 10 dvhlmod φ U LMod
17 5 7 15 16 ldualvbase φ Base D = F
18 13 14 17 3sstr4d φ C Base D
19 eqid Scalar U = Scalar U
20 eqid 0 Scalar U = 0 Scalar U
21 eqid Base U = Base U
22 19 20 21 5 lfl0f U LMod Base U × 0 Scalar U F
23 16 22 syl φ Base U × 0 Scalar U F
24 1 3 2 21 10 dochoc1 φ ˙ ˙ Base U = Base U
25 eqidd φ Base U × 0 Scalar U = Base U × 0 Scalar U
26 19 20 21 5 6 lkr0f U LMod Base U × 0 Scalar U F L Base U × 0 Scalar U = Base U Base U × 0 Scalar U = Base U × 0 Scalar U
27 16 23 26 syl2anc φ L Base U × 0 Scalar U = Base U Base U × 0 Scalar U = Base U × 0 Scalar U
28 25 27 mpbird φ L Base U × 0 Scalar U = Base U
29 28 fveq2d φ ˙ L Base U × 0 Scalar U = ˙ Base U
30 29 fveq2d φ ˙ ˙ L Base U × 0 Scalar U = ˙ ˙ Base U
31 24 30 28 3eqtr4d φ ˙ ˙ L Base U × 0 Scalar U = L Base U × 0 Scalar U
32 eqid 0 U = 0 U
33 1 3 2 21 32 doch1 K HL W H ˙ Base U = 0 U
34 10 33 syl φ ˙ Base U = 0 U
35 29 34 eqtrd φ ˙ L Base U × 0 Scalar U = 0 U
36 32 4 lss0ss U LMod R S 0 U R
37 16 11 36 syl2anc φ 0 U R
38 35 37 eqsstrd φ ˙ L Base U × 0 Scalar U R
39 9 lcfls1lem Base U × 0 Scalar U C Base U × 0 Scalar U F ˙ ˙ L Base U × 0 Scalar U = L Base U × 0 Scalar U ˙ L Base U × 0 Scalar U R
40 23 31 38 39 syl3anbrc φ Base U × 0 Scalar U C
41 40 ne0d φ C
42 eqid Base Scalar U = Base Scalar U
43 eqid D = D
44 10 adantr φ x Base Scalar D a C b C K HL W H
45 11 adantr φ x Base Scalar D a C b C R S
46 simpr3 φ x Base Scalar D a C b C b C
47 eqid + D = + D
48 simpr2 φ x Base Scalar D a C b C a C
49 simpr1 φ x Base Scalar D a C b C x Base Scalar D
50 eqid Scalar D = Scalar D
51 eqid Base Scalar D = Base Scalar D
52 19 42 7 50 51 16 ldualsbase φ Base Scalar D = Base Scalar U
53 52 adantr φ x Base Scalar D a C b C Base Scalar D = Base Scalar U
54 49 53 eleqtrd φ x Base Scalar D a C b C x Base Scalar U
55 1 2 3 4 5 6 7 19 42 43 9 44 45 48 54 lclkrslem1 φ x Base Scalar D a C b C x D a C
56 1 2 3 4 5 6 7 19 42 43 9 44 45 46 47 55 lclkrslem2 φ x Base Scalar D a C b C x D a + D b C
57 56 ralrimivvva φ x Base Scalar D a C b C x D a + D b C
58 50 51 15 47 43 8 islss C T C Base D C x Base Scalar D a C b C x D a + D b C
59 18 41 57 58 syl3anbrc φ C T