Show that the norms $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are both continuous. 
Show that the norms  $\|\cdot\|_1 : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ and $\|\cdot\|_\infty : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ are both continuous.

For the first one if $x \in \mathbb{R}^n$, then $\|x\|_1=|x_1|+|x_2|+\dots+ |x_n|$ this seems just to be a sum of multiple projections e.g $\operatorname{pr}_j: \mathbb{R}^n \to \mathbb{R}$, where $j  \in \{0, \dots n\}$. So as the sum of continuous functions this is continuous? For the second one I'm not sure how to approach. I know that if I have for example $\max\{f,g\}$ then this equals $\frac12(f+g)+\frac12|f-g|$ so do I have that $$\max\{|x_1|,|x_2|, \dots|x_n|\}= \frac1n(|x_1|+\dots|x_n|)+\frac1n||x_1|-|x_2|-\dots-|x_n||$$ and I could show that this is continuous by the fact that all $x_i$ are continous since the previous result?
 A: A easy counter example: $\max\{1,2,3\} = 3$, but
$$ \dfrac{1}{3}\big(1+2+3\big) + \dfrac{1}{3}\big| 1-2-3 \big| = 2+\dfrac{4}{3} = 3\dfrac{1}{3} \ne 3, $$
Hint:
(I guess the norm $d_2$ means $\|\cdot\|_2$.)
You can prove the following results in turn:

*

*The finite sum of continuous functions is continuous, And all $|x_j|$ are continuous.

*$\max\{a,b\}=\dfrac{1}{2}(a+b)+\dfrac{1}{2}|a-b|$.

*$\max\{a,b,c\}=\max\big\{\max\{a,b\},c\big\}$.

A: Every norm is continuous w.r.t topology generated by this norm because it's Lipschitz function with Lipschitz constant $L = 1$, and since all norms are equivalent on the finite dimenshion space we get the statement.
A: It suffices to show that there are constants $C,K > 0$ such that
$$
\|x\|_1 \le C \|x\|_2  \quad \text{and} \quad \|x\|_{\infty} \le K \|x\|_2.
$$
This is because if $(x_n)$ is a sequence with $\|x_n - x\|_2 \to 0$ then $\|x_n -x\|_1 \le C \|x_n - x\|_2 \to 0$, and similarly for the other norm. Hence, the sequential characterisation of continuity implies that $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are continuous with respect to the Euclidean distance on $\mathbb{R}^n$.
For the first inequality we can use the Cauchy-Schwarz inequality to get
$$
\|x\|_1 = \sum_{i=1}^n |x_i| =  \sum_{i=1}^n 1 \cdot |x_i| \le \Big(\sum_{i=1}^n 1^2\Big)^{1/2}\Big(\sum_{i=1}^n |x_i|^2\Big)^{1/2} = \sqrt{n} \, \|x\|_2.
$$
For the second inequality
$$
\|x\|_{\infty} = \max_{1 \le i \le n} |x_i| = \sqrt{\max_{1 \le i \le n} |x_i|^2} \le \sqrt{\sum_{i=1}^n |x_i|^2} = \|x\|_2.$$
