Prove verification:$ \|x\|_{\infty}$, $\|x\|_{1}$, $\|f\|_{\infty}$ are norms 
Prove The Following Are Norms:
Let $V=\mathbb{R}^n$ or  $V=\mathbb{C}^n$. For $x\in V$ define
  $$
\|x\|_{\infty}=\max\{|x_{k}|:k=1,2,...,n\},
\quad \|x\|_{1}=\sum_{k=1}^{n}|x_{k}|
$$
Let $V=C[a,b]$. For $f\in V$ define
  $$
\|f\|_{\infty}=\max\{|f(x)|:a\leq x\leq b\}
$$

Proof: 
1a. $$\max\{|x_{k}|:k=1,2,...,n\}\geq 0$$ as $|x_{k}|\geq 0$ for all $k$ and for all $k$
$$
\max\{|x_{k}|:k=1,2,...,n\}= 0 \iff |x_{k}|=0
$$ 
1b.\begin{align}\|\alpha x\|_{\infty}&=\max\{|\alpha x_{k}|:k=1,2,...,n\} \\
&\stackrel{(1)}{=}\max\{|\alpha|\cdot| x_{k}|:k=1,2,...,n\} \\
&=|a|\cdot \max\{| x_{k}|:k=1,2,...,n\} \\
&=|a|\|x\|_{\infty}
\end{align} 
$(1)$ $|\alpha x_{k}|=|a||x_k|$ for all $k$
1c.\begin{align}\|x+y\|_{\infty}&=\max\{|x_{k}+y_{k}|:k=1,2,...,n\} \\
&\stackrel{(1)}{\leq}\max\{|x_{k}|+|y_{k}|:k=1,2,...,n\} \\
&\stackrel{(2)}{\leq} \max\{|x_{k}|:k=1,2,...,n\}+\max\{|y_{k}|:k=1,2,...,n\} \\
&=\|x\|_\infty+\|y\|_\infty
\end{align}
$(1)$ $|x_k+y_k|=|x_k|+|y_k|$
$(2)$ Which property of $\max$ allow us to say that $\max\{|x|+|y|\}=\max\{|x|\}+\max\{|y|\}$?
2a.$$\|x\|_{1}=\sum_{k=1}^{n}|x_{k}|$$ for all k $|x_{k}|\geq 0$ and $|x_{k}|=0 \iff x_k=0$
2b. $$\|\alpha \cdot x\|_{1}=\sum_{k=1}^{n}|\alpha \cdot x_{k}|\stackrel{(1)}{=}\sum_{k=1}^{n}|\alpha| \cdot |x_{k}|=|\alpha|\sum_{k=1}^{n} |x_{k}|=|\alpha|\|x\|_{1}$$
$(1)$ $|\alpha \cdot x_{k}|=|\alpha|\cdot|x_{k}|$
2c.  $$\|x+y\|_{1}=\sum_{k=1}^{n}|x_{k}+y_{k}|\leq \sum_{k=1}^{n}(|x_{k}|+|y_{k}|)\leq  \sum_{k=1}^{n}|x_{k}|+ \sum_{k=1}^{n}|y_{k}|=\|x\|_{1}+\|y\|_{1}$$
3a.$$\|f\|_{\infty}=\max\{|f(x)|:a\leq x\leq b\}$$ for all $x\in[a,b]$ $|f(x)|\geq 0$ and $|f(x)|=0 \iff f(x)=0$ therefore
$$
\max\{|f(x)|:a\leq x\leq b\}\geq 0$$ and $$\max\{|f(x)|:a\leq x\leq b\}=0\iff f(x)=0
$$
3b. \begin{align}\|\alpha\cdot f\|_{\infty}&=\max\{|\alpha f(x)|:a\leq x\leq b\}\\
&\stackrel{(1)}{=}\max\{|\alpha|\cdot |f(x)|:a\leq x\leq b\}\\
&=|\alpha|\cdot \max\{|f(x)|:a\leq x\leq b\} \\
&= |\alpha|\|f\|_\infty
\end{align}
$(1)$ $|\alpha f(x)|=|\alpha||f(x)|$
3c. \begin{align}\|f+g\|_{\infty}&\stackrel{(1)}{\leq} \max\{|f(x)|+|g(x)|:a\leq x\leq b\}\\
&\stackrel{(2)}{\leq} \max\{|f(x)|:a\leq x\leq b\}+\max\{|g(x)|:a\leq x\leq b\} \\
&=\|f\|_{\infty}+\|g\|_{\infty}\end{align}
$(1)$ $|f(x)+g(x)|\leq|f(x)|+|g(x)|$
 A: Too many questions. Let's take a look at the first one.

Proof: 
  1a. (Mathematical proofs consists of logical sentences. Beginning a sentence with a formula confuses the reader.) Note that
  $$
\max\{|x_{k}|:k=1,2,...,n\}\geq 0$$ as $|x_{k}|\geq 0$ for all $k$. and for all $k$ Thus (well, I don't see how your first sentence relates to the following one, which is true though),
  $$
\max\{|x_{k}|:k=1,2,...,n\}= 0 \iff |x_{k}|=0
$$ 
1b.\begin{align}\|\alpha x\|_{\infty}&=\max\{|\alpha x_{k}|:k=1,2,...,n\} \\
&\stackrel{(1)}{=}\max\{|\alpha|\cdot| x_{k}|:k=1,2,...,n\} \\
&=|a|\cdot \max\{| x_{k}|:k=1,2,...,n\} \\
&=|a|\|x\|_{\infty}
\end{align} 
$(1)$ $|\alpha x_{k}|=|a||x_k|$ for all $k$ (the next equality deserves more explanation.)
1c.\begin{align}\|x+y\|_{\infty}&=\max\{|x_{k}+y_{k}|:k=1,2,...,n\} \\
&\stackrel{(1)}{\leq}\max\{|x_{k}|+|y_{k}|:k=1,2,...,n\} \\
&\stackrel{(2)}{\leq} \max\{|x_{k}|:k=1,2,...,n\}+\max\{|y_{k}|:k=1,2,...,n\} \\
&=\|x\|_\infty+\|y\|_\infty
\end{align}
$(1)$ $|x_k+y_k|=|x_k|+|y_k|$ (Oops.)
$(2)$ Which property of $\max$ allow us to say that $\max\{|x|+|y|\}=\max\{|x|\}+\max\{|y|\}$? (This expression is incorrect. If you don't know how to prove (2), it is a good question to ask. Try it!)

