Proving a Function is a Norm

So we seek to show the following:

Theorem 1.1   The function from $\mathbb{R}^{n}$ into $\mathbb{R}$ defined by

$$\Vert \mathbf{x}\Vert _{2} = (\sum_{j=1}^{n} x_{j}^{2})^{1/2}$$

is a norm.

Proof. Let $\mathbf{x}\in \mathbb{R}^{n}$. We prove the proprieties in order.

1. Since $x^{2}_{j} \geq 0$ for $j=1,2,\ldots,n$. Then
   $$\sum_{j=1}^{n} x_{j}^{2} \geq 0$$
   The positive branch of the square root function is an increasing function so we have
   $$(\sum_{j=1}^{n} x_{j}^{2})^{1/2} \geq 0.$$

2. Note that $\mathbf{x}= \mathbf{0}$ is equivalent to $x_{j} = 0$ for $j=1,2,\ldots,n$. Of course, this quickly implies
   $$(\sum_{j=1}^{n} x_{j}^{2})^{1/2} = 0.$$
   There is a subtle point here. We also need to show
   $$(\sum_{j=1}^{n} x_{j}^{2})^{1/2} = 0.$$
   implies that $x_{j} = 0$ for $j=1,2,\ldots,n$. There are many reasonable candidates for norms that fail this property. It is often easier to prove the contrapositive. Recall the contrapositive of ($A$ implies $B$) is ($\neg B$ implies $\neg A$). These expressions are logically equivalent. Prove one and you have proved the other. So we assume $\mathbf{x}\neq \mathbf{0}$. This means there is an index $k$ such that $x_{k} \neq 0$. Since $f(x) = x^{2}$ has only one root at $x = 0$, then we know $x_{k}^{2} > 0$. Furthermore $x_{j}^{2} \geq 0$ for the remaining components of $\mathbf{x}$. Hence,
   $$(\sum_{j=1}^{n} x_{j}^{2})^{1/2} > 0.$$
   Using the increasing property of the square root function, we get $\Vert \mathbf{x}\Vert _{2} > 0$. (This may seems like too many words to show something that is obvious, but this obvious property fails enough mathematicians had to create the concept of semi-norm).

3. Using the fact that $\sqrt{\alpha^{2}} = \vert\alpha\vert$, quickly get this property.

4. The triangle inequality is also the problematic part of showing a function is a norm. Let $\mathbf{x},\mathbf{y}\in \mathbb{R}^{n}$. Then
   $$\begin{split} \Vert \mathbf{x}+ \mathbf{y}\Vert^{2}_{2} &= (... ...thbf{x}\Vert _{2} + \Vert \mathbf{y}\Vert _{2})^{2} \end{split}$$

$\qedsymbol$