absolute value

Absolute value theorems PhL 8.14.09 ************** A. Theorems where a and b are complex numbers *********************** _______________________________________________________________ Theorem 1: |ab| = |a||b| Proof: Let a = Aeiα and b = Beiβ. Then |ab| = AB, |a| = A, |b| = B, QED. You can think of a and b as vectors in the space Cn where n = 1. If we made n-dimensional vectors A = (a1, a2. ,,,,, an) and similarly for b, then this Theorem 1 is seen to be the famous Cauchy Schwarz Inequality |AB| ≤ |A| |B| in Cn. So let's write this out as a variant of Theorem 1: [ note that ai and bi are just arbitrary complex numbers ] _______________________________________________________________ Theorem 1A: |a1b1 + a2b2| ≤ |a1+a2| |b1+ b2| ≤ (|a1| + |a2|) (|b1| + |b2|) | Σi aibi | ≤ |Σi ai| | Σi bi | ≤ (Σi | ai|) (Σi | bi|) Proof: The first ≤ sign on both lines is the CSI as just noted (and which I will prove in a moment). The second ≤ sign on each line comes from Theorem 2 (or Theorem 2C) stated below. According to Ahlfors, it is the comparison of the first with the third expressions which he calls Cauchy's Inequality on p 10. But others refer to the CSI itself as Cauchy's Inequality. _______________________________________________________________ Theorem 1B: |<x,y>| ≤ <x,x><y,y> x,y any Hilbert Space, including Cn The Cauchy-Schwarz Inequality = CSI in my notes (apologies to the Russian B) Proof: Stakgold page 107 has a short proof, but it is specific to Rn. Wiki gives a short proof which works for any Hilbert Space, as follows. (The trick is to pick λ in a smart way.) Remember that an inner product is a complex number and that <a,b> = <b,a>*. This proof uses <x,λy> = <x,y> which is the Stakgold/Math form, not the Quantum Mechanics form. Here is some algebra to go with the above: (x,x) -(x,y) - λ (y,x) + |λ|2 (y,y) = = (x,x) -(y,x)(x,y)/(y,y) - (x,y) (y,x)/(y,y) + (x,y) (y,x)/ (y,y) // two terms cancel = (x,x) -(y,x)(x,y)/(y,y) = (x,x) - |(x,y)|2/(y,y) ≥ 0 So you conclude that |(x,y)|2≤ (x,x) (y,y) ______________________________________________________________ Theorem 2: |a + b| ≤ |a| + |b| Proof # 1. Let a = x+iy and b = u + iv. Then |a + b|2 = |(x+u) + i(y+v)|2 = (x+u)2 + (y+v)2 |a|2 = x2 + y2 |b|2 = u2 + v2 So we need to show that, where all square roots are positive, [(x+u)2 + (y+v)2]1/2 ≤ [x2 + y2]1/2 + [u2 + v2]1/2 (5) We will now construct a sequence of steps which, when run backwards, show the above is true. Square both sides: (x+u)2 + (y+v)2 ≤ x2 + y2 + u2 + v2 + 2 [x2 + y2]1/2[u2 + v2]1/2 (4) 2xu + 2yv ≤ 2 [x2 + y2]1/2[u2 + v2]1/2 xu + yv ≤ [x2 + y2]1/2[u2 + v2]1/2 (3) Square both sides x2u2 + y2v2 ≤ [x2 + y2][u2 + v2] = x2u2 + y2 v2 + y2u2 + x2 v2 (2) 0 ≤ y2u2 + x2 v2 (1) So start with (1) and go backwards and you end up with 5 and hence theorem is proved. Proof #2. Think of a and b as 2D vectors in the complex plane: Then |a + b| ≤ |a| + |b| because the length of side a+b is obviously at most the sum of the lengths of sides a and b and this happens when θ = π. Hence this theorem is always called "the triangle inequality". Comment: we know that c2 ≡ (a+b)2 = a2 + b2 + 2ab = a2 + b2 – 2abcosθ where now a and b are lengths of triangle sides, not complex numbers. Thus, the RHS here ranges from (a-b)2 to (a+b)2 . So for sure this tells us c2 ≤ (a+b)2 or c ≤ a+b, which is our theorem's claim. ______________________________________________________________ Theorem 2A: |a – b| ≤ |a| + |b| Proof: Just take b → -b in Theorem 2, QED. Just a different triangle: ______________________________________________________________ Theorem 2B: |a-b| ≤ |a-c| + |b-c|. Proof: start with Theorem 2A which says |A - B| ≤ |A| + |B|. Then set A = a-c and B = b-c, QED. Just another triangle: ______________________________________________________________ Theorem 2C: | a + b + c| ≤ |a| + |b| + |c| and so on | Σi ai| ≤ Σi | ai| Proof: Let a+b = d and apply theorem 2 twice. Can install any signs you want on the LHS. ******************* B. Theorems where a and b are real numbers: ********************* Since real numbers are a subset of complex numbers, all versions of Theorem 1 and Theorem 2 given above are true for reals. But there are now more theorems for reals only that we can produce. ______________________________________________________________ Theorem 3: |a-b| ≥ |a| - |b| a,b real Proof: Again for |b| ≥ |a| the RHS is negative and LHS is positive, so theorem is blatantly true, so we only need consider |a| > |b|. In this realm, there are two cases to consider: a and b have the same sign, or they have opposite sign. This seems to involve 4 different cases, but the ab = ++ case is the same as the ab = - - case, since nothing on either side of our theorem changes under this double sign change. Similarly, the ab = -+ case is the same as the ab = +- case. So we can now examine our only two unique cases: ++ a-b ≥ a-b true +- Here, let B = - b > 0. Then theorem says a+B ≥ a - B which is obviously true Thus our theorem is proved. ______________________________________________________________ Theorem 4: |a-b| ≥ | |a| - |b| | a,b real Proof: If |a| > |b|, we showed this to be true in Theorem 3. If |b| ≥ |a|, Theorem 3 would also tell us that |a-b| ≥ |b| - |a| , just by swapping a ↔ b. Thus, if |b| ≥ |a|, we know that |a-b| ≥ | |a| - |b| |. Thus we have shown that |a-b| ≥ | |a| - |b| | in both cases |a| > |b| and |b| ≥ |a|, QED. This is a stronger theorem in a sense than Theorem. ______________________________________________________________ Theorem 5: |a-b| < ε => |a| - |b| < ε a,b real Proof: By theorem 3 we have |a-b| ≥ |a| - |b|. Adding the hypothesis of Theorem 5 we have |a| - |b| ≤ |a-b| < ε Therefore |a| - |b| < ε, QED. ______________________________________________________________ Here is some verification of some of our theorems from http://en.wikipedia.org/wiki/Absolute_value, these involve only reals: