P-Norm is Norm/Complex Numbers

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

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

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:

Thus is satisfied.

Suppose that $\lambda \in K$.

Thus is satisfied.

If $\mathbf x = \sequence 0$ and $\mathbf y = \sequence 0$, then by we have equality.

If $\mathbf x + \mathbf y = \sequence 0$ and both $\bf x$ and $\bf y$ nonvanishing, then by we get a strict inequality.

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

Thus is satisfied.

All norm axioms are seen to be satisfied.

Hence the result.