## Subsection1.2.4The vector $p$-norms

A vector norm is a measure of the magnitude of a vector. The Euclidean norm (length) is merely the best known such measure. There are others. A simple alternative is the 1-norm.

###### Definition1.2.4.1. Vector 1-norm.

The vector 1-norm, $\| \cdot \|_1 : \C^m \rightarrow \mathbb R \text{,}$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_1 = \vert \chi_0 \vert + \vert \chi_1 \vert + \cdots + \vert \chi_{m-1} \vert = \sum_{i=0}^{m-1} \vert \chi_i \vert. \end{equation*}
###### Homework1.2.4.1.

Prove that the vector 1-norm is a norm.

Solution

We show that the three conditions are met:

Let $x, y \in \C^m$ and $\alpha \in \mathbb C$ be arbitrarily chosen. Then

• $x \neq 0 \Rightarrow \| x \|_1 > 0$ ($\| \cdot \|_1$ is positive definite):

Notice that $x \neq 0$ means that at least one of its components is nonzero. Let's assume that $\chi_j \neq 0 \text{.}$ Then

\begin{equation*} \| x \|_1 = \vert \chi_0 \vert + \cdots + \vert \chi_{m-1} \vert \geq \vert \chi_j \vert > 0 . \end{equation*}
• $\| \alpha x \|_1 = \vert \alpha \vert \| x \|_1$ ($\| \cdot \|_1$ is homogeneous):

\begin{equation*} \begin{array}{l} \| \alpha x \|_1 ~~~=~~~~ \lt \mbox{ scaling~a~vector-scales-its-components; definition} \gt \\ \vert \alpha \chi_0 \vert + \cdots + \vert \alpha \chi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ algebra} \gt \\ \vert \alpha \vert \vert \chi_0 \vert + \cdots + \vert \alpha \vert \vert \chi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ algebra} \gt \\ \vert \alpha \vert ( \vert \chi_0 \vert + \cdots + \vert \chi_{m-1} \vert ) \\ ~~~=~~~~ \lt \mbox{ definition} \gt \\ \vert \alpha \vert \| x \|_1. \end{array} \end{equation*}
• $\| x + y \|_1 \leq \| x \|_1 + \| y \|_1$ ($\| \cdot \|_1$ obeys the triangle inequality):

\begin{equation*} \begin{array}{l} \| x + y \|_1 \\ ~~~=~~~~ \lt \mbox{ vector~addition;~definition~of~1-norm} \gt \\ \vert \chi_0 + \psi_0 \vert + \vert \chi_1 + \psi_1 \vert + \cdots + \vert \chi_{m-1} + \psi_{m-1} \vert \\ ~~~\leq~~~~ \lt \mbox{ algebra} \gt \\ \vert \chi_0 \vert + \vert \psi_0 \vert + \vert \chi_1 \vert + \vert \psi_1 \vert + \cdots + \vert \chi_{m-1} \vert + \vert \psi_{m-1} \vert \\ ~~~=~~~~ \lt \mbox{ commutivity} \gt \\ \vert \chi_0 \vert + \vert \chi_1 \vert + \cdots + \vert \chi_{m-1} \vert + \vert \psi_0 \vert + \vert \psi_1 \vert + \cdots + \vert \psi_{m-1} \vert \\ ~~~= ~~~~ \lt \mbox{ associativity;~definition} \gt \\ \| x \|_1 + \| y \|_1. \end{array} \end{equation*}

The vector 1-norm is sometimes referred to as the "taxi-cab norm". It is the distance that a taxi travels, from one point on a street to another such point, along the streets of a city that has square city blocks.

Another alternative is the infinity norm.

###### Definition1.2.4.2. Vector $\infty$-norm.

The vector $\infty$-norm, $\| \cdot \|_\infty : \C^m \rightarrow \mathbb R \text{,}$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_\infty = \max( \vert \chi_0 \vert, \ldots , \vert \chi_{m-1} ) = \max_{i=0}^{m-1} \vert \chi_i \vert. \end{equation*}

The infinity norm simply measures how large the vector is by the magnitude of its largest entry.

###### Homework1.2.4.2.

Prove that the vector $\infty$-norm is a norm.

Solution

We show that the three conditions are met:

Let $x, y \in \C^m$ and $\alpha \in \mathbb C$ be arbitrarily chosen. Then

• $x \neq 0 \Rightarrow \| x \|_\infty > 0$ ($\| \cdot \|_\infty$ is positive definite):

Notice that $x \neq 0$ means that at least one of its components is nonzero. Let's assume that $\chi_j \neq 0 \text{.}$ Then

\begin{equation*} \| x \|_\infty = \max_{i=0}^{m-1} \vert \chi_i \vert \ge \vert \chi_j \vert > 0. \end{equation*}
• $\| \alpha x \|_\infty = \vert \alpha \vert \| x \|_\infty$ ($\| \cdot \|_\infty$ is homogeneous):

\begin{equation*} \begin{array}{lcl} \| \alpha x \|_\infty \amp =\amp \max_{i=0}^{m-1} \vert \alpha \chi_i \vert \\ \amp =\amp \max_{i=0}^{m-1} \vert \alpha \vert \vert \chi_i \vert \\ \amp =\amp \vert \alpha \vert \max_{i=0}^{m-1} \vert \chi_i \vert \\ \amp =\amp \vert \alpha \vert \| x \|_\infty. \end{array} \end{equation*}
• $\| x + y \|_\infty \leq \| x \|_\infty + \| y \|_\infty$ ($\| \cdot \|_\infty$ obeys the triangle inequality):

\begin{equation*} \begin{array}{lcl} \| x + y \|_\infty \amp =\amp \max_{i=0}^{m-1} \vert \chi_i + \psi_i \vert \\ \amp \leq\amp \max_{i=0}^{m-1} ( \vert \chi_i \vert + \vert \psi_i \vert ) \\ \amp \leq\amp \max_{i=0}^{m-1} \vert \chi_i \vert + \max_{i=0}^{m-1} \vert \psi_i \vert \\ \amp = \amp \| x \|_\infty + \| y \|_\infty. \end{array} \end{equation*}

In this course, we will primarily use the vector 1-norm, 2-norm, and $\infty$-norms. For completeness, we briefly discuss their generalization: the vector $p$-norm.

###### Definition1.2.4.3. Vector $p$-norm.

Given $p \geq 1 \text{,}$ the vector $p$-norm $\| \cdot \|_p : \C^m \rightarrow \mathbb R$ is defined for $x \in \C^m$ by

\begin{equation*} \| x \|_p = \sqrt[p]{\vert \chi_0 \vert^p + \cdots + \vert \chi_{m-1} \vert^p} = \left( \sum_{i=0}^{m-1} \vert \chi_i \vert^p \right)^{1/p}. \end{equation*}

The proof of this result is very similar to the proof of the fact that the 2-norm is a norm. It depends on Hölder's inequality, which is a generalization of the Cauchy-Schwartz inequality:

We skip the proof of Hölder's inequality and Theorem 1.2.4.4. You can easily find proofs for these results, should you be interested.

###### Remark1.2.4.6.

The vector 1-norm and 2-norm are obviously special cases of the vector $p$-norm. It can be easily shown that the vector $\infty$-norm is also related:

\begin{equation*} \lim_{p \rightarrow \infty} \| x \|_p = \| x \|_{\infty}. \end{equation*}
###### Ponder This1.2.4.3.
Consider Homework 1.2.3.3. Try to elegantly formulate this question in the most general way you can think of. How do you prove the result?
###### Ponder This1.2.4.4.

Consider the vector norm $\| \cdot \|: \Cm \rightarrow \mathbb R \text{,}$ the matrix $A \in \mathbb C^{m \times n}$ and the function $f: \Cn \rightarrow \mathbb R$ defined by $f( x ) = \| A x \| \text{.}$ For what matrices $A$ is the function $f$ a norm?