:: HOLDER_1 semantic presentation  Show TPTP formulae Show IDV graph for whole article:: Showing IDV graph ... (Click the Palm Trees again to close it)

registration
let x be real number ;
cluster right_closed_halfline x -> non empty ;
coherence
not right_closed_halfline x is empty
proof end;
end;

theorem :: HOLDER_1:1  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q being Real st 0 < p & 0 < q holds
for a being Real st 0 <= a holds
(a to_power p) * (a to_power q) = a to_power (p + q)
proof end;

theorem :: HOLDER_1:2  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q being Real st 0 < p & 0 < q holds
for a being Real st 0 <= a holds
(a to_power p) to_power q = a to_power (p * q)
proof end;

theorem Th3: :: HOLDER_1:3  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p being Real st 0 < p holds
for a, b being Real st 0 <= a & a <= b holds
a to_power p <= b to_power p
proof end;

theorem Th4: :: HOLDER_1:4  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q, a, b being Real st 1 < p & (1 / p) + (1 / q) = 1 & 0 < a & 0 < b holds
( a * b <= ((a #R p) / p) + ((b #R q) / q) & ( a * b = ((a #R p) / p) + ((b #R q) / q) implies a #R p = b #R q ) & ( a #R p = b #R q implies a * b = ((a #R p) / p) + ((b #R q) / q) ) )
proof end;

theorem Th5: :: HOLDER_1:5  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q, a, b being Real st 1 < p & (1 / p) + (1 / q) = 1 & 0 <= a & 0 <= b holds
( a * b <= ((a to_power p) / p) + ((b to_power q) / q) & ( a * b = ((a to_power p) / p) + ((b to_power q) / q) implies a to_power p = b to_power q ) & ( a to_power p = b to_power q implies a * b = ((a to_power p) / p) + ((b to_power q) / q) ) )
proof end;

Lm1: for a being Real_Sequence st ( for n being Nat holds 0 <= a . n ) holds
for n being Nat holds a . n <= (Partial_Sums a) . n
proof end;

Lm2: for a being Real_Sequence st ( for n being Nat holds 0 <= a . n ) holds
for n being Nat holds 0 <= (Partial_Sums a) . n
proof end;

Lm3: for a being Real_Sequence st ( for n being Nat holds 0 <= a . n ) holds
for n being Nat st (Partial_Sums a) . n = 0 holds
for k being Nat st k <= n holds
a . k = 0
proof end;

Lm4: for a being Real_Sequence
for n being Nat st ( for k being Nat st k <= n holds
a . k = 0 ) holds
(Partial_Sums a) . n = 0
proof end;

theorem Th6: :: HOLDER_1:6  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q being Real st 1 < p & (1 / p) + (1 / q) = 1 holds
for a, b, ap, bq, ab being Real_Sequence st ( for n being Nat holds
( ap . n = (abs (a . n)) to_power p & bq . n = (abs (b . n)) to_power q & ab . n = abs ((a . n) * (b . n)) ) ) holds
for n being Nat holds (Partial_Sums ab) . n <= (((Partial_Sums ap) . n) to_power (1 / p)) * (((Partial_Sums bq) . n) to_power (1 / q))
proof end;

theorem Th7: :: HOLDER_1:7  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p being Real st 1 < p holds
for a, b, ap, bp, ab being Real_Sequence st ( for n being Nat holds
( ap . n = (abs (a . n)) to_power p & bp . n = (abs (b . n)) to_power p & ab . n = (abs ((a . n) + (b . n))) to_power p ) ) holds
for n being Nat holds ((Partial_Sums ab) . n) to_power (1 / p) <= (((Partial_Sums ap) . n) to_power (1 / p)) + (((Partial_Sums bp) . n) to_power (1 / p))
proof end;

theorem Th8: :: HOLDER_1:8  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a, b being Real_Sequence st ( for n being Nat holds a . n <= b . n ) & b is convergent & a is non-decreasing holds
( a is convergent & lim a <= lim b )
proof end;

theorem Th9: :: HOLDER_1:9  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a, b, c being Real_Sequence st ( for n being Nat holds a . n <= (b . n) + (c . n) ) & b is convergent & c is convergent & a is non-decreasing holds
( a is convergent & lim a <= (lim b) + (lim c) )
proof end;

theorem Th10: :: HOLDER_1:10  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p being Real st 0 < p holds
for a, ap being Real_Sequence st a is convergent & ( for n being Nat holds 0 <= a . n ) & ( for n being Nat holds ap . n = (a . n) to_power p ) holds
( ap is convergent & lim ap = (lim a) to_power p )
proof end;

theorem :: HOLDER_1:11  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p being Real st 0 < p holds
for a, ap being Real_Sequence st a is summable & ( for n being Nat holds 0 <= a . n ) & ( for n being Nat holds ap . n = ((Partial_Sums a) . n) to_power p ) holds
( ap is convergent & lim ap = (Sum a) to_power p & ap is non-decreasing & ( for n being Nat holds ap . n <= (Sum a) to_power p ) )
proof end;

theorem :: HOLDER_1:12  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p, q being Real st 1 < p & (1 / p) + (1 / q) = 1 holds
for a, b, ap, bq, ab being Real_Sequence st ( for n being Nat holds
( ap . n = (abs (a . n)) to_power p & bq . n = (abs (b . n)) to_power q & ab . n = abs ((a . n) * (b . n)) ) ) & ap is summable & bq is summable holds
( ab is summable & Sum ab <= ((Sum ap) to_power (1 / p)) * ((Sum bq) to_power (1 / q)) )
proof end;

theorem :: HOLDER_1:13  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for p being Real st 1 < p holds
for a, b, ap, bp, ab being Real_Sequence st ( for n being Nat holds
( ap . n = (abs (a . n)) to_power p & bp . n = (abs (b . n)) to_power p & ab . n = (abs ((a . n) + (b . n))) to_power p ) ) & ap is summable & bp is summable holds
( ab is summable & (Sum ab) to_power (1 / p) <= ((Sum ap) to_power (1 / p)) + ((Sum bp) to_power (1 / p)) )
proof end;