Cauchy-Schwarz inequality, Triangle inequality

(d) Cauchy-Schwarz inequality - Triangle inequality

Theorem :

Let V be an inner product space over F. Then for all x, y Є V and c Є F, the following statements are true.

(a) ||cx|| = |c|.||x||

(b) || x || = 0 if and only if x = 0. In any case, ||x|| ≥ 0

(c) (Cauchy-Schwarz Inequality) |(x,y)| = |x|| . || y ||

(d) (Triangle Inequality) || x + y || ≤ ||x|| + || y ||

Proof: Let V be an inner product space over F.

Let x,y Є V and c Є F

(c) If y = 0, then the result is true.

So assume that y ≠ 0.

For any c Є F, we have

In particular,

(d) We have,

where Re<x, y> denotes the real part of the complex number <x, y>.

=> ||x + y|| ≤ || x || + || y ||

Example :

For Fⁿ, we may apply Cauchy-Schwarz and triangle inequality to the standard inner product to obtain the following well-known inequalities :

Problem 1.

In C([0, 1]), let f(t) = t and g(t) = e^t verify Cauchy-Schwarz and triangle inequality.

Solution :

(a) Cauchy-Schwarz inequality states that

.'. from (2) & (5), we get

(b) Triangle inequality states that

Problem 2.

Let x = (2,1 + i, i) and y = (2-i, 2, 1+2i) be vectors in C³. Verify Cauchy-Schwarz and triangle inequality.

Solution :

(a) Cauchy-Schwarz inequality states that

(b) Triangle inequality states that

From (5), (6), we get

|| x + y || ≤ || x || + || y ||

Problem 3.

Let V be an inner product space, and suppose that x and y are orthogonal vectors in V. Prove that || x + y ||² = || x ||² + || y ||². Deduce the Pythagorean theorem in R².

Solution :

x and y are orthogonal => <x,y> = 0

Let a right angled triangle with perpendicular sides of length || x || and || y ||, let length of base is ||x|| and length of perpendicular is || y ||.

By the geometry of a right angled triangle and using the concept of a norm, the third side that is the hypotenuse of the triangle can be given by x + y and the length of the hypotenuse is ||x + y||

.'. (1) ⇒ Hypotenuse² = base² + Perpendicular²

which is the Pythagorean Theorem in R²

Problem 4.

Prove the parallelogram law on an inner product space V, that is, show that 

What does this equation state about parallelograms in R²?

Solution :

Consider a parallelogram in R² whose adjacent sides of length |x|| and ||y||, then from the geometry of the parallelogram, ||x + y|| and ||x - y|| denotes the length of two diagonals of the parallelogram.

Hence, the equation (1) states that the sum of the area of the squares made on the diagonal of a parallelogram is equal to twice of the sum of the area of the squares made on the adjacent sides of the given parallelogram.

Problem 5.

Let V be an inner product space. Prove that

 where Re<x,y> denotes the real part of the complex number <x,y>

Solution:

Let V be an inner product space.

Similarly,

Problem 6.

Let V be  an inner product space. Prove that

Solution :

By using the traingle inequality,

Problem 7.

Let V be an inner product space over F. Prove the polar identities: For all x, y Є V.

Solution :