Metamath Proof Explorer


Theorem fourierdlem59

Description: The derivative of H is continuous on the given interval. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses fourierdlem59.f φF:
fourierdlem59.x φX
fourierdlem59.a φA
fourierdlem59.b φB
fourierdlem59.n0 φ¬0AB
fourierdlem59.fdv φFX+AX+B:X+AX+Bcn
fourierdlem59.c φC
fourierdlem59.h H=sABFX+sCs
Assertion fourierdlem59 φH:ABcn

Proof

Step Hyp Ref Expression
1 fourierdlem59.f φF:
2 fourierdlem59.x φX
3 fourierdlem59.a φA
4 fourierdlem59.b φB
5 fourierdlem59.n0 φ¬0AB
6 fourierdlem59.fdv φFX+AX+B:X+AX+Bcn
7 fourierdlem59.c φC
8 fourierdlem59.h H=sABFX+sCs
9 1 adantr φsABF:
10 2 adantr φsABX
11 elioore sABs
12 11 adantl φsABs
13 10 12 readdcld φsABX+s
14 9 13 ffvelcdmd φsABFX+s
15 7 adantr φsABC
16 14 15 resubcld φsABFX+sC
17 eqcom s=00=s
18 17 biimpi s=00=s
19 18 adantl sABs=00=s
20 simpl sABs=0sAB
21 19 20 eqeltrd sABs=00AB
22 21 adantll φsABs=00AB
23 5 ad2antrr φsABs=0¬0AB
24 22 23 pm2.65da φsAB¬s=0
25 24 neqned φsABs0
26 16 12 25 redivcld φsABFX+sCs
27 26 8 fmptd φH:AB
28 ioossre AB
29 28 a1i φAB
30 dvfre H:ABABH:domH
31 27 29 30 syl2anc φH:domH
32 ovex ABV
33 32 a1i φABV
34 eqidd φsABFX+sC=sABFX+sC
35 eqidd φsABs=sABs
36 33 16 12 34 35 offval2 φsABFX+sC÷fsABs=sABFX+sCs
37 8 36 eqtr4id φH=sABFX+sC÷fsABs
38 37 oveq2d φDH=DsABFX+sC÷fsABs
39 reelprrecn
40 39 a1i φ
41 16 recnd φsABFX+sC
42 eqid sABFX+sC=sABFX+sC
43 41 42 fmptd φsABFX+sC:AB
44 12 recnd φsABs
45 eldifsn s0ss0
46 44 25 45 sylanbrc φsABs0
47 eqid sABs=sABs
48 46 47 fmptd φsABs:AB0
49 eqidd φsABFX+s=sABFX+s
50 eqidd φsABC=sABC
51 33 14 15 49 50 offval2 φsABFX+sfsABC=sABFX+sC
52 51 eqcomd φsABFX+sC=sABFX+sfsABC
53 52 oveq2d φdsABFX+sCds=DsABFX+sfsABC
54 14 recnd φsABFX+s
55 eqid sABFX+s=sABFX+s
56 54 55 fmptd φsABFX+s:AB
57 15 recnd φsABC
58 eqid sABC=sABC
59 57 58 fmptd φsABC:AB
60 eqid DFX+AX+B=DFX+AX+B
61 cncff FX+AX+B:X+AX+BcnFX+AX+B:X+AX+B
62 6 61 syl φFX+AX+B:X+AX+B
63 1 2 3 4 60 62 fourierdlem28 φdsABFX+sds=sABFX+AX+BX+s
64 ioosscn X+AX+B
65 64 a1i φX+AX+B
66 ax-resscn
67 66 a1i φ
68 62 67 fssd φFX+AX+B:X+AX+B
69 ssid
70 69 a1i φ
71 cncfcdm FX+AX+B:X+AX+BcnFX+AX+B:X+AX+BcnFX+AX+B:X+AX+B
72 70 6 71 syl2anc φFX+AX+B:X+AX+BcnFX+AX+B:X+AX+B
73 68 72 mpbird φFX+AX+B:X+AX+Bcn
74 ioosscn AB
75 74 a1i φAB
76 2 recnd φX
77 2 3 readdcld φX+A
78 77 rexrd φX+A*
79 78 adantr φsABX+A*
80 2 4 readdcld φX+B
81 80 rexrd φX+B*
82 81 adantr φsABX+B*
83 3 adantr φsABA
84 83 rexrd φsABA*
85 4 rexrd φB*
86 85 adantr φsABB*
87 simpr φsABsAB
88 ioogtlb A*B*sABA<s
89 84 86 87 88 syl3anc φsABA<s
90 83 12 10 89 ltadd2dd φsABX+A<X+s
91 4 adantr φsABB
92 iooltub A*B*sABs<B
93 84 86 87 92 syl3anc φsABs<B
94 12 91 10 93 ltadd2dd φsABX+s<X+B
95 79 82 13 90 94 eliood φsABX+sX+AX+B
96 65 73 75 76 95 fourierdlem23 φsABFX+AX+BX+s:ABcn
97 63 96 eqeltrd φdsABFX+sds:ABcn
98 iooretop ABtopGenran.
99 eqid TopOpenfld=TopOpenfld
100 99 tgioo2 topGenran.=TopOpenfld𝑡
101 98 100 eleqtri ABTopOpenfld𝑡
102 101 a1i φABTopOpenfld𝑡
103 7 recnd φC
104 40 102 103 dvmptconst φdsABCds=sAB0
105 0cnd φ0
106 75 105 70 constcncfg φsAB0:ABcn
107 104 106 eqeltrd φdsABCds:ABcn
108 40 56 59 97 107 dvsubcncf φsABFX+sfsABC:ABcn
109 53 108 eqeltrd φdsABFX+sCds:ABcn
110 40 102 dvmptidg φdsABsds=sAB1
111 1cnd φ1
112 75 111 70 constcncfg φsAB1:ABcn
113 110 112 eqeltrd φdsABsds:ABcn
114 40 43 48 109 113 dvdivcncf φsABFX+sC÷fsABs:ABcn
115 38 114 eqeltrd φH:ABcn
116 cncff H:ABcnH:AB
117 fdm H:ABdomH=AB
118 115 116 117 3syl φdomH=AB
119 118 feq2d φH:domHH:AB
120 31 119 mpbid φH:AB
121 cncfcdm H:ABcnH:ABcnH:AB
122 67 115 121 syl2anc φH:ABcnH:AB
123 120 122 mpbird φH:ABcn