Multivariable Advanced Calculus

56 BASIC LINEAR ALGEBRALemma 3.8.9 If a, b ≥ 0 and p ′ is defined by 1 p + 1 p ′= 1, thenab ≤ app + bp′p ′ .Proof of the Proposition: If x or y equals the zero vector there is nothing toprove. Therefore, assume they are both nonzero. Let A = ( ∑ ni=1 |x i| p ) 1/p and B =( ∑ni=1 |y i| p′) 1/p ′ . Then using Lemma 3.8.9,n∑i=1[|x i | |y i |nA B ≤ ∑ 1pi=1= 1 p1A pi=1( |xi |An ∑= 1 p + 1 p ′ = 1) p+ 1 ( ) p′]|yi |p ′ B|x i | p + 1 p ′ 1B pn ∑i=1|y i | p′and so(n∑n) 1/p (∑n) 1/p′∑|x i | |y i | ≤ AB = |x i | p |y i | p′ .i=1This proves the proposition. i=1Theorem 3.8.10 The p norms do indeed satisfy the axioms of a norm.Proof: It is obvious that ||·|| pdoes indeed satisfy most of the norm axioms. Theonly one that is not clear is the triangle inequality. To save notation write ||·|| in placeof ||·|| pin what follows. Note also that p p= p − 1. Then using the Holder inequality,′||x + y|| p =≤=≤n∑|x i + y i | pi=1n∑|x i + y i | p−1 |x i | +i=1n∑|x i + y i | p p ′ |x i | +i=1i=1i=1n∑|x i + y i | p−1 |y i |i=1n∑|x i + y i | p p ′ |y i |i=1(∑ n) 1/p′⎡( n) 1/p (∑n) ⎤ 1/p∑|x i + y i | p ⎣ |x i | p + |y i | p ⎦i=1= ||x + y|| p/p′ ( ||x|| p+ ||y|| p)i=1so dividing by ||x + y|| p/p′ , it follows||x + y|| p ||x + y|| −p/p′ = ||x + y|| ≤ ||x|| p+ ||y|| p() )p − p p= p(1 − 1 ′p= p 1 ′ p = 1. . This proves the theorem. It only remains to prove Lemma 3.8.9.