P-Norm is Norm/P-Sequence Space

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Theorem

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


Proof


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Norm Axiom $\text N 1$: Positive Definiteness

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:

\(\ds \norm {\mathbf x}_p\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 0}^\infty \size {x_n}^p\) \(=\) \(\ds 0\) 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
\(\ds \leadsto \ \ \) \(\ds \bf x\) \(=\) \(\ds \sequence 0_{n \mathop \in \N}\)

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

$\Box$


Norm Axiom $\text N 2$: Positive Homogeneity

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

\(\ds \norm {\alpha \cdot \mathbf x}_p\) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty \size {\alpha x_n}^p }^{\frac 1 p}\)
\(\ds \) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty \size \alpha^p \size {x_n}^p }^{\frac 1 p}\)
\(\ds \) \(=\) \(\ds \size {\alpha} \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p }^{\frac 1 p}\)
\(\ds \) \(=\) \(\ds \size {\alpha} \norm {\mathbf x}_p\)

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

$\Box$


Norm Axiom $\text N 3$: Triangle Inequality

From P-Norm is Norm/Complex Numbers 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:

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

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

$\blacksquare$


Sources

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