# Semi-norms and symmetric bilinear forms

Let $V$ be a real vector space and $\cdot : V^2 \to \mathbb{R}$ a symmetric bilinear form. In this post I will first show that Cauchy-Schwarz inequality is equivalent to $\cdot$ being semi-definite. I will then show that for a non-negative positive-homogeneous function $f : V \to \mathbb{R}$ the triangle inequality, convexity, and quasi-convexity are all equivalent. Finally, I will show that if  $|\cdot|: V \to \mathbb{R} : |x| = \sqrt{|x \cdot x|}$, then $|\cdot|$  is a semi-norm if and only if $\cdot$ is semi-definite.

## Definitions

A bilinear form $\cdot : V^2 \to \mathbb{R}$ is called

• symmetric, if $\forall x, y \in V: x \cdot y = y \cdot x$,
• positive-semi-definite, if $\forall x \in V: x\cdot x \geq 0$,
• negative-semi-definite, if $\forall x \in V: x \cdot x \leq 0$,
• semi-definite, if it is either positive-semi-definite or negative-semi-definite,
• indefinite, if it is not semi-definite, and
• fulfilling the Cauchy-Schwarz inequality, if $\forall x, y \in V: (x \cdot y)^2 \leq (x\cdot x)(y \cdot y).$

A function $f : V \to \mathbb{R}$ is called

• positive-homogeneous, if $\forall x \in V: \forall \alpha \in \mathbb{R}: f(\alpha x) = |\alpha| f(x)$,
• fulfilling the triangle inequality, if $\forall x, y \in V: f(x + y) \leq f(x) + f(y),$
• convex if $\forall x, y \in V: \forall t \in [0, 1] \subset \mathbb{R}: f((1 - t)x + ty) \leq (1 - t)f(x) + tf(y)$, and
• quasi-convex, if $\forall x, y \in V: \forall t \in [0, 1] \subset \mathbb{R}: f((1 - t)x + ty) \leq \max\{f(x), f(y)\}.$

## Cauchy-Schwarz inequality is equivalent to semi-definiteness

Assume the Cauchy-Schwarz inequality holds but that $\cdot$ is indefinite. Then there exists $x, y \in V$ such that $x \cdot x > 0$ and $y \cdot y < 0$. Then $(x \cdot y)^2 \leq (x \cdot x)(y \cdot y)$ does not hold since the left-hand side is non-negative and the right-hand side is negative. This is a contradiction. Therefore $\cdot$ is semi-definite.

Assume $\cdot$ is semi-definite. First, if $y \cdot y = 0$, then the Cauchy-Schwarz inequality holds trivially. Assume $y \cdot y \neq 0$. Decompose $x$ as $x = x_{\parallel} + x_{\perp}$, where $x_{\parallel} = \frac{x \cdot y}{y \cdot y} y$ is the orthogonal projection of $x$ to $y$, and $x_{\perp} = x - x_{\parallel}$ is the rejection of $x$ from $y$. Then, by semi-definiteness, the Cauchy-Schwarz inequality states that

Again by semi-definiteness, and $x_{\parallel} \cdot x_{\perp} = 0$,

Therefore the Cauchy-Schwarz inequality holds. QED.

## Convexity implies quasi-convexity

Let $f : V \to \mathbb{R}$ be a convex function. Let $x, y \in V$, and $t \in [0, 1] \subset \mathbb{R}$. Now

Therefore $f$ is quasi-convex. QED.

## Quasi-convexity + non-negativity + positive-homogenuity implies triangle inequality

Let $f : V \to \mathbb{R}$ be a non-negative quasi-convex positive-homogeneous function. For $x, y \in V$, let $t = f(y) / (f(x) + f(y))$. Now

Using positive-homogenuity and non-negativeness,

Therefore $f$ fulfills the triangle inequality. QED.

## For positive-homogeneous functions convexity and triangle inequality are equivalent

Let $f : V \to \mathbb{R}$ be a positive-homogeneous function. Assume $f$ is convex. Then $f(x + y) = 2f(0.5 x + 0.5 y) \leq f(x) + f(y).$ Therefore $f$ fulfills the triangle inequality. Assume $f$ fulfills the triangle inequality. Let $t \in [0, 1] \in \mathbb{R}$. Then $f((1 - t)x + ty) \leq f((1 - t)x) + f(ty) = (1 - t)f(x) + tf(y)$. Therefore $f$ is convex. QED.

## Semi-definiteness is equivalent to triangle inequality

Let $\cdot : V^2 \to \mathbb{R}$ be a symmetric bilinear form, and $|\cdot|: V \to \mathbb{R} : |x| = \sqrt{|x \cdot x|}$.  Assume $\cdot$ is indefinite. Then by indefiniteness there exists $x, y \in V$ such that $x \cdot x > 0$ and $y \cdot y < 0$. Now one can solve the quadratic equation $|(1 - t)x + ty| = 0$ for $t \in \mathbb{R}$. The solution is

The discriminant is always positive, since $(x \cdot x)(y \cdot y) < 0$. Therefore, there are two points $a, b \in V$, with $|a| = 0$ and $|b| = 0$, which lie on the same line as $x$ and $y$. Either $x$ or $y$ is a convex combination of $a$ and $b$. Without loss of generality, assume it is $x$. If $|\cdot|$ were convex, it would hold that $|x| = 0$. Since this is not the case, $|\cdot|$ is not convex and the triangle inequality does not hold.

Assume $\cdot$ is semi-definite. Then the Cauchy-Schwarz inequality holds and $(x \cdot x)(y \cdot y) \geq 0$. Now,

Thus the triangle inequality holds. QED.

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