13.03.2012 (25.05.2015)

Let be a real vector space and a symmetric bilinear form. In this post I will first show that Cauchy-Schwarz inequality is equivalent to being semi-definite. I will then show that for a non-negative positive-homogeneous function the triangle inequality, convexity, and quasi-convexity are all equivalent. Finally, I will show that if is such that . then is a semi-norm if and only if is semi-definite.

A bilinear form is called

*symmetric*, if ,*positive-semi-definite*, if ,*negative-semi-definite*, if ,*semi-definite*, if ,*indefinite*, if it is not semi-definite, and*fulfilling the Cauchy-Schwarz inequality*, if ,*degenerate*, if there exists such that , for all ,

for all .

A function is called

*positive-homogeneous*, if ,*fulfilling the triangle inequality*, if ,*convex*if , and*quasi-convex*, if .

for all , , and .

Assume is indefinite. Then there exists such that and . It follows that does not hold; the left-hand side is non-negative and the right-hand side is negative. Therefore does not fulfill the Cauchy-Schwarz inequality.

Assume is semi-definite.

Let . Then

It follows that

Similarly,

shows that

Therefore,

Since is arbitrary, we have that . Therefore the Cauchy-Schwarz inequality holds.

Decompose as , where is the orthogonal projection of to , and is the rejection of from . By semi-definiteness, and ,

This is the Cauchy-Schwarz inequality.

Let be a convex function. Then

for all , and . Therefore is quasi-convex.

Let be a non-negative quasi-convex positive-homogeneous function. For , let . Now

Using positive-homogenuity and non-negativeness,

Therefore fulfills the triangle inequality.

Let be a positive-homogeneous function. Assume is convex. Then

for all . Therefore fulfills the triangle inequality. Assume fulfills the triangle inequality. Then

for all . Therefore f is convex.

Let be a symmetric bilinear form, and be such that .

Assume is indefinite. By indefiniteness, there exists such that and . Now one can solve the quadratic equation for . The solution is

The discriminant is always positive, since . Therefore, there are two points , with and , which lie on the same line as and . Either or is a convex combination of and . Without loss of generality, assume it is . If were convex, it would hold that . Since this is not the case, is not convex and the triangle inequality does not hold.

Assume is semi-definite. Then the Cauchy-Schwarz inequality holds and . Now,

Thus the triangle inequality holds.