Euclidean space geometry: Cauchy-Schwarz inequality
At first glance, the scalar product has nothing to do with the distance notion. It has, and the Cauchy-Schwarz inequality provides one of the links.
Statement (Cauchy-Schwarz inequality) (I) For any vectors one has .
(II) If the inequality sign turns into equality, , then is proportional to : .
Proof. (I) If at least one of the vectors is zero, both sides of the inequality are 0 and there is nothing to prove. To exclude the trivial case, suppose that none of is zero and, therefore, are positive. Consider a real-valued function of a real number defined by . Here we have a norm of a linear combination .
We see that is a parabola with branches looking upward (because the senior coefficient is positive). By nonnegativity of the squared norm, and the parabola lies above the horizontal axis in the plane. Hence, the quadratic equation may have at most one real root. This means that the discriminant of the equation is non-positive: Applying square roots to both sides of we finish the proof of the first part.
(II) In case of the equality sign the discriminant is 0. Therefore the parabola touches the horizontal axis where . But we know that this implies which is just another way of writing .
Do you think this proof is tricky? During the long history of development of mathematics, mathematicians have invented many tricks, small and large. No matter how smart you are, you cannot reinvent all of them. Studying the existing tricks is a necessity. By the way, the definition of what is tricky and what is not is strictly personal and time-dependent.
Consequences of the Cauchy-Schwarz inequality
Exercise 1. Prove the triangle inequality.
Proof. From the expression for the norm of a linear combination
Definition. For any two nonzero vectors from the Cauchy-Schwarz inequality we have Therefore we can define the cosine of the angle between by
This definition agrees with the definition of orthogonality: means that the angle between is and should be zero.
Playing with balls
Once we have a norm, we can define the distance between by For any and let us put This is a set of points whose distance from is less than Therefore it is a ball centered at and with radius
Exercise 2. (Application of the triangle inequality) Consider two balls and where so the first ball is larger. Under what condition the smaller ball is contained in the larger ball:
Solution. There are two possible ways to solve this exercise. One is to look at the geometry first and then try to translate it to math. The other is just try to see the solution from calculations. We follow the second way.
The inclusion relationship in terms of sets (1) is equivalent to a point-wise statement: any element of the smaller ball belongs to the larger ball. This means that we have to start with the assumption and arrive to Using the triangle inequality we have
We want to be smaller than This is achieved if we require
Exercise 3. A set is called open if any its element belongs to together with some ball Prove that is open.
Proof. Take any We have to produce such From the previous exercise we have this inclusion if .