Let $K \in \C^d$, 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: {{begin-eqn}} {{eqn | l = \norm {\mathbf x}_p       | r = 0 }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^d \size {x_n}^p       | r = 0       | c = raising to power $p > 0$ }} {{eqn | ll= \leadsto       | l = \size {x_n}       | r = 0       | c = Sum of Non-Negatives vanishes iff Summands vanish }} {{eqn | ll= \leadsto       | l = x_n       | r = 0       | c = Complex Modulus equals Zero iff Zero }} {{eqn | ll= \leadsto       | l = \bf x       | r = \sequence 0_{n \mathop \in \N, \, n \mathop \le d} }} {{end-eqn}} Thus norm axiom $(\text N 1)$ is satisfied. {{qed|lemma}} === Norm Axiom $(\text N 2)$ === Suppose, $\lambda \in K$. {{begin-eqn}} {{eqn | l = \norm {\lambda \mathbf x}_p       | r = \paren {\sum_{n \mathop = 0}^d \size {\lambda x_n}^p}^{1/p} }} {{eqn | r = \paren {\size{\lambda}^p \sum_{n \mathop = 0}^d \size {x_n}^p}^{1/p} }} {{eqn | r = \size {\lambda} \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1/p} }} {{eqn | r = \size {\lambda} \norm {\mathbf x}_p }} {{end-eqn}} Thus norm axiom $(\text N 2)$ is satisfied. {{qed|lemma}} === 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 $p$: {{begin-eqn}} {{eqn | l = \norm {\bf x + \bf y}_p^p       | r = \sum_{n \mathop = 0}^d \size {x_n + y_n} \size {x_n + y_n}^{p \mathop - 1} }} {{eqn | o = \le       | r = \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}       | c = Triangle Inequality for Complex Numbers }} {{eqn | o = \le       | r = \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} }       | c = Modulus of Product }} {{eqn | o = \le       | r = \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       | c = Hölder's Inequality for Sums: $\dfrac 1 p + \dfrac 1 q = 1$ }} {{eqn | o = \le       | r = \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}       | c = Transformation of $p$-Norm: $q \paren {p - 1} = p$ }} {{eqn | ll= \leadsto       | l = \norm {\mathbf x + \mathbf y}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1}       | o = \le       | r = \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} }} {{eqn | ll= \leadsto       | l = \norm {\bf x + \bf y}_p       | o = \le       | r = \norm {\bf x}_p  + \norm {\bf y}_p        | c = Division by $\norm {\bf x + \bf y}_p^{p \mathop - 1}$ }} {{end-eqn}} Thus norm axiom $(\text N 3)$ is satisfied. {{qed|lemma}} All norm axioms are seen to be satisfied. Hence the result. {{qed}}