# P-Norm is Norm/Complex Numbers

## Theorem

The $p$-norm on the complex numbers is a norm.

## Proof

Let $K \in \C^d$, where $d \in \N_{>0}$.

### Norm Axiom $\text N 1$: Positive Definiteness

Suppose $\sequence {x_n}_{n \mathop \in \set {1, 2, \ldots, d} } \in K$.

By definition of $p$-norm:

$\ds \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1 / p}$

The complex modulus of $x_n$ is real and non-negative.

We have the results:

Sum of Non-Negative Reals is Non-Negative
Power of Positive Real Number is Positive
Zero Raised to Positive Power is Zero

Hence:

$\norm {\mathbf x}_p \ge 0$

Suppose that $\norm {\mathbf x}_p = 0$.

Then:

 $\ds \norm {\mathbf x}_p$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sum_{n \mathop = 0}^d \size {x_n}^p$ $=$ $\ds 0$ raising to power $p > 0$ $\ds \leadsto \ \$ $\ds \size {x_n}$ $=$ $\ds 0$ Sum of Non-Negatives vanishes iff Summands vanish $\ds \leadsto \ \$ $\ds x_n$ $=$ $\ds 0$ Complex Modulus equals Zero iff Zero $\ds \leadsto \ \$ $\ds \bf x$ $=$ $\ds \sequence 0_{n \mathop \in \N, \, n \mathop \le d}$

Thus Norm Axiom $\text N 1$: Positive Definiteness is satisfied.

$\Box$

### Norm Axiom $\text N 2$: Positive Homogeneity

Suppose that $\lambda \in K$.

 $\ds \norm {\lambda \mathbf x}_p$ $=$ $\ds \paren {\sum_{n \mathop = 0}^d \size {\lambda x_n}^p}^{1 / p}$ $\ds$ $=$ $\ds \paren {\size {\lambda}^p \sum_{n \mathop = 0}^d \size {x_n}^p}^{1 / p}$ $\ds$ $=$ $\ds \size {\lambda} \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1 / p}$ $\ds$ $=$ $\ds \size {\lambda} \norm {\mathbf x}_p$

Thus Norm Axiom $\text N 2$: Positive Homogeneity is satisfied.

$\Box$

### Norm Axiom $\text N 3$: Triangle Inequality

If $\mathbf x = \sequence 0$ and $\mathbf y = \sequence 0$, then by Norm Axiom $\text N 1$: Positive Definiteness we have equality.

If $\mathbf x + \mathbf y = \sequence 0$ and both $\bf x$ and $\bf y$ nonvanishing, then by Norm Axiom $\text N 1$: Positive Definiteness we get a strict inequality.

If $\mathbf x + \mathbf y \ne \sequence 0$, then consider p-norm raised to the power of $p$:

 $\ds \norm {\bf x + \bf y}_p^p$ $=$ $\ds \sum_{n \mathop = 0}^d \size {x_n + y_n} \size {x_n + y_n}^{p \mathop - 1}$ $\ds$ $\le$ $\ds \sum_{n \mathop = 0}^d \size {x_n} \size {x_n + y_n}^{p \mathop - 1} + \sum_{n \mathop = 0}^d \size {y_n} \size {x_n + y_n}^{p \mathop - 1}$ Triangle Inequality for Complex Numbers $\ds$ $\le$ $\ds \sum_{n \mathop = 0}^d \size {x_n \paren{x_n + y_n}^{p \mathop - 1} } + \sum_{n \mathop = 0}^d \size {y_n \paren{ x_n + y_n}^{p \mathop - 1} }$ Modulus of Product $\ds$ $\le$ $\ds \norm {\bf x}_p \norm {\paren{\mathbf x + \mathbf y}^{p \mathop - 1} }_q + \norm {\mathbf y}_p \norm {\paren{\mathbf x + \mathbf y}^{p \mathop - 1} }_q$ Hölder's Inequality for Sums: $\dfrac 1 p + \dfrac 1 q = 1$ $\ds$ $\le$ $\ds \norm {\bf x}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1} + \norm {\mathbf y}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1}$ Transformation of $p$-Norm: $q \paren {p - 1} = p$ $\ds \leadsto \ \$ $\ds \norm {\mathbf x + \mathbf y}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1}$ $\le$ $\ds \norm {\mathbf x}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1} + \norm {\bf y}_p \norm {\bf x + \bf y}_p^{p \mathop - 1}$ $\ds \leadsto \ \$ $\ds \norm {\bf x + \bf y}_p$ $\le$ $\ds \norm {\bf x}_p + \norm {\bf y}_p$ Division by $\norm {\bf x + \bf y}_p^{p \mathop - 1}$

Thus Norm Axiom $\text N 3$: Triangle Inequality is satisfied.

$\Box$

All norm axioms are seen to be satisfied.

Hence the result.

$\blacksquare$