P-Norm is Norm

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

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

Norm Axiom $(\text N 1)$
Suppose $\sequence {x_n}_{n \mathop \in {\N}, \, n \mathop \le d} \in K$.

By definition of $p$-norm:


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

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, $\norm {\mathbf x}_p = 0$.

Then:

Thus norm axiom $(\text N 1)$ is satisfied.

Norm Axiom $(\text N 2)$
Suppose, $\lambda \in K$.

Thus norm axiom $(\text N 2)$ is satisfied.

Norm Axiom $(\text N 3)$
If $\mathbf x = \sequence 0$ and $\mathbf y = \sequence 0$, then by $\paren {\text N 1}$ we have equality.

If $\mathbf x + \mathbf y = \sequence 0$ and both $\bf x$ and $\bf y$ nonvanishing, then by $\paren {\text N 1}$ we get a strict inequality.

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

Thus norm axiom $(\text N 3)$ is satisfied.

All norm axioms are seen to be satisfied.

Hence the result.