P-Norm is Norm/Complex Numbers

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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.

$\blacksquare$


Sources

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