Multivariable Advanced Calculus

54 BASIC LINEAR ALGEBRATheorem 3.8.4 (Cauchy Schwarz)Let H be an inner product space. The followinginequality holds for x and y ∈ H.|(x · y)| ≤ (x · x) 1/2 (y · y) 1/2 (3.33)Equality holds in this inequality if and only if one vector is a multiple of the other.Proof: Let θ ∈ F such that |θ| = 1 andθ (x · y) = |(x · y)|Consider p (t) ≡ ( x + θty · x + tθy ) where t ∈ R. Then from the above list of propertiesof the dot product,0 ≤ p (t) = (x · x) + tθ (x · y) + tθ (y · x) + t 2 (y · y)= (x · x) + tθ (x · y) + tθ(x · y) + t 2 (y · y)= (x · x) + 2t Re (θ (x · y)) + t 2 (y · y)= (x · x) + 2t |(x · y)| + t 2 (y · y) (3.34)and this must hold for all t ∈ R. Therefore, if (y · y) = 0 it must be the case that|(x · y)| = 0 also since otherwise the above inequality would be violated. Therefore, inthis case,|(x · y)| ≤ (x · x) 1/2 (y · y) 1/2 .On the other hand, if (y · y) ≠ 0, then p (t) ≥ 0 for all t means the graph of y = p (t) isa parabola which opens up and it either has exactly one real zero in the case its vertextouches the t axis or it has no real zeros. From the quadratic formula this happensexactly when4 |(x · y)| 2 − 4 (x · x) (y · y) ≤ 0which is equivalent to 3.33.It is clear from a computation that if one vector is a scalar multiple of the other thatequality holds in 3.33. Conversely, suppose equality does hold. Then this is equivalentto saying 4 |(x · y)| 2 −4 (x · x) (y · y) = 0 and so from the quadratic formula, there existsone real zero to p (t) = 0. Call it t 0 . Thenp (t 0 ) ≡ ( x + θt 0 y, x + t 0 θy ) = ∣ ∣x + θty ∣ ∣ 2 = 0and so x = −θt 0 y. This proves the theorem. Note that in establishing the inequality, I only used part of the above properties ofthe dot product. It was not necessary to use the one which says that if (x · x) = 0 thenx = 0.Now the length of a vector can be defined.Definition 3.8.5 Let z ∈ H. Then |z| ≡ (z · z) 1/2 .Theorem 3.8.6 For length defined in Definition 3.8.5, the following hold.|z| ≥ 0 and |z| = 0 if and only if z = 0 (3.35)If α is a scalar, |αz| = |α| |z| (3.36)|z + w| ≤ |z| + |w| . (3.37)