P-Norm is Norm/P-Sequence Space

Theorem
The $p$-norm on the $p$-sequence space is a norm.

Norm Axiom $(\text N 1)$
Let $x \in \ell^p$ with $1 \le p < \infty$.

By definition of $p$-norm:


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

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 $\alpha \in \set {\R, \C}$.

Norm Axiom $(\text N 3)$
From P-Norm is Norm we have that:


 * $\ds \paren {\sum_{n \mathop = 0}^d \size {x_n + y_n}^p}^{\frac 1 p} \le \paren {\sum_{n \mathop = 0}^d \size {x_n}^p }^{\frac 1 p} + \paren {\sum_{n \mathop = 0}^d \size {y_n}^p }^{\frac 1 p}$

$\map f z = z^{\frac 1 p}$ is a continuous function for $z \ge 0$ and $p > 0$.

For $\mathbf x \in \ell^p$, changing $d$ is equivalent to changing $z$ in the interval $0 \le z < \infty$.

Take the composite limit $d \to \infty$.

Then:


 * $\norm {\mathbf x + \mathbf y}_p \le \norm {\mathbf x}_p + \norm {\mathbf y}_p$