Let V be a vector space with inner product <x,y> and norm ||x|| = <x,x>^1/2.
Prove the Cauchy-Schwarz inequality <x,y> <= ||x|| ||y||.
Hint given in book: If x,y != 0, set c = 1/||x|| and d = 1/||y|| and use the fact that
||cx ± dy|| >= 0.
Here, the inner product is not necessarily the dot product, nor the norm the Euclidean norm. All that is necessary is that the two fulfill properties of inner products and norms.
For inner products, the given properties are
1. <x,y> = <y,x>
2. <x + y, z> = <x,y> + <x,z>
3. <cx,y> = c<x,y> = <x,cy>
4. <x,x> > 0 if x != 0.
For norms, the given properties are
1. ||x|| > 0 if x != 0
2. ||cx|| = |c|||x||
3. ||x + y|| <= ||x|| + ||y||
The inequality ||cx ± dy|| >= 0 is given as true.
The Attempt at a Solution
I have attempted to solve the problem by starting with the Cauchy-Schwarz inequality and working towards the inequality given in the hint as true:
<x,y> >= ||x|| ||y||
-2<x,y> >= -2||x|| ||y||
-2cd<x,y> >= -2
1 - 2cd<x,y> + 1 >= 0
(c^2)||x||^2 - 2cd<x,y> + (d^2)||d||^2 >= 0
(c^2)<x,x> -cd<x,y> - cd<y,x> + (d^2)<y,y> >= 0
<cx,cx> + <cx, -dy> + <-dy, cx> + <dy,dy> >= 0
<cx - dy, cx - dy> >= 0
||cx - dy||^2 >= 0
Taking the square root here yields
||cx - dy|| >= 0 or ||cx - dy|| <= 0
but the norm of a vector is a non-negative number, so the right inequality simplifies to
||cx - dy|| = 0
The two inequalities can then be combined to yield
||cx - dy|| >= 0, which was given as a fact at the beginning of the problem.
Do you see any flaws in my logic, or any steps that could be simplified or added to make the proof more clear?