Question about sup norm Let $x \in \mathbb{R}^n$. Define $|x| = \max\{ |x_1|,...,|x_n|\} $. I want to show that this is a norm on $R^n$. This is my reasoning. First, notice
$$ |x| = \max\{ |x_i| \} \geq |x_i| \; \forall i ,\;\; |x_i| \geq 0 \implies |x| \geq 0$$
$$ |cx| = \max\{ |cx_i| \} = \max\{ |c||x_i| \} = |c| \max\{ |x_i| \} = |c| |x|$$
$$ |x + y| = \max\{ |x_i + y_i| \} = |x_i +y_i| \; \text{for some i} \leq |x_i| + |y_i|  \leq \max\{|x_i|\} + \max\{|y_i|\} = |x| + |y| $$ 
Hence, this is indeed a norm on $R^n$. Is this correct? Thanks in advance for your help.
Also, I have a bit of trouble seeing if this inequality is true:
$$ \max\{ |x_1|,...,|x_n|\} \leq \sqrt{ \sum x_i^2} \leq \sqrt{n} \max\{ |x_1|,...,|x_n|\} $$
Can I show this by induction? Thanks
 A: We need to show that
$$ \max\big\{ |x_1|,...,|x_n|\big\} \,\,\stackrel{(1)}{\leq}\,\, \sqrt{ \sum x_i^2} \,\,\stackrel{(2)}{\leq}\,\, \sqrt{n} \max\big\{ |x_1|,...,|x_n|\big\} $$
For $(1)$, let $|x_j|=\max\big\{ |x_1|,...,|x_n|\big\}$, for some $j=1,\ldots,n$. Then
$$
|x_j|^2\le |x_1|^2+\cdots+|x_n|^2,
$$
and hence
$$
\max\big\{ |x_1|,...,|x_n|\big\}=|x_j|\le \sqrt{|x_1|^2+\cdots+|x_n|^2}.
$$
For $(2)$, if once again $|x_j|=\max\big\{ |x_1|,...,|x_n|\big\}$, then
$$
|x_1|^2+\cdots+|x_n|^2\le \underbrace{|x_j|^2+|x_j^2|+\cdots+|x_j|^2}_{n\,\,\,\text{times}}=n|x_j|^2,
$$
and thus
$$
\sqrt{|x_1|^2+\cdots+|x_n|^2}\le \sqrt{n\,|x_j|^2}=\sqrt{n}\,|x_j|=\sqrt{n}\,\max\big\{ |x_1|,...,|x_n|\big\}.
$$
A: Your proof for the triangular inequality is correct. As for the second inequality, let $(x_{1},\ldots,x_{n}) \in \mathbb{R}^{n}$. There is no need to prove your inequality by induction. A direct proof works fine. I will first prove 
$$ \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} \leq \sqrt{\sum_{i=1}^{n} x_{i}^{2}} \tag{1}$$
and then :
$$ \sqrt{\sum_{i=1}^{n} x_{i}^{2}} \leq \sqrt{n} \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} \tag{2}$$
For $(1)$, notice that $\max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} = \vert x_{i_{0}} \vert$ for some $i_{0} \in \left\{1, \ldots, n \right\}$. Moreover, 
$$
\vert x_{i_{0}} \vert = \sqrt{x_{i_{0}}^{2}} \leq \sqrt{\sum_{i=1}^{n} x_{i}^{2}}$$ because $\displaystyle 0 \leq x_{i_{0}}^{2} \leq \sum_{i=1}^{n} x_{i}^{2} $ and the function $x \, \longmapsto \, \sqrt{x}$ is increasing on $\mathbb{R}^{+}$. That gives you $(1)$.
For $(2)$, by definition, we have : $\forall i \in \left\{ 1,\ldots,n \right\}, \; \vert x_{i} \vert \leq \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\}$. Squaring this last inequality gives :
$$ \forall i \in \left\{ 1,\ldots,n \right\}, \; x_{i}^{2} = \vert x_{i} \vert^{2} \leq \big( \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} \big)^{2} \tag{$\star$}$$
Then, sum $(\star)$ for $i$ from $1$ to $n$ and you will get :
$$ 0 \leq \sum_{i=1}^{n} x_{i}^{2} \leq n \big( \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} \big)^{2} $$
Again, $x \, \longmapsto \, \sqrt{x}$ is increasing on $\mathbb{R}^{+}$ and this gives you $(2)$ :
$$ \sqrt{\sum_{i=1}^{n} x_{i}^{2}} \leq \sqrt{n} \max \left\{ \vert x_{1} \vert, \ldots, \vert x_{n} \vert \right\} $$
