Does this make sense what I have written? Prooving that $\{|x|,|y|\}$ is a norm
I tried to proove the following in the last post
Show that the norm $||(x,y)||=max\{|x|,|y|\}$ where ||.|| is the norm and $(x,y) \in \mathbb{R}^{2}$ really is a norm by verifiying all the Axioms.
But they baisically told me that it is a horrible mess and my tutor is complaing generally about the clearaty of my Proofs. Is this version better?
Let $v$ denote a vector $\in \mathbb{R}^2$ and $x$,$y$ are the two components.
First we need to show that$||v||=0$ $\iff$ $v=0$
We use $A \Rightarrow B \land B \Rightarrow A \Rightarrow (A \iff B)$ in order to do this
Consider first $\Rightarrow$:
We have $||v||=||(x,y)||=max\{|x|,|y|\}=0$ This implies that $|x| \leq 0$ and $|y| \leq 0$ which implies since $0 \leq |x| \leq 0$ and $0 \leq |y| \leq 0$ that $x=0$ and $y=0$. Thus we have $v=0$ where $v=(x,y)$.
the other direction $ \Leftarrow$
Let $v=0$ then $x=0$ $y=0$. We find that $||v||=0$ since the maximum of two zeros is oviously zero.
The second we need to show is that $||\lambda v||=\lambda ||v||$ where $\lambda$ is a real number.
We now try to show that $||\lambda v||=|\lambda| ||v||$
We have that $max\{|\lambda x|,|\lambda y|\}=max\{|\lambda||y|,|\lambda||y|\}$
Let wlog $|\lambda||x|$ be the maximum of $|\lambda||x|$ and $|\lambda||y|$ meaning $|x||\lambda|\leq |x||\lambda|$ and $|y||\lambda|\leq |\lambda||x|$. It is easy to see that by dividing by $\lambda$ we have $|x|\leq |x|$ and $|y| \leq |x|$ thus we find $max\{|x|,|y|\}=|x|$. But then we have $||v||=max\{|\lambda x|, |\lambda y|\}=max\{|\lambda||x|,|\lambda||y|\}=|\lambda||x|=|\lambda|max\{|x|,|y|\}$ and finally we can conclude that $||\lambda v||=|\lambda|||v||$
The third and last Axiom we need to show is that $||v+w||\leq||v||+||w||$ where $w$ simply denotes another vector. And let $v=(x_1,y_1)$ and let $w=(x_2,y_2)$.
Let wlog be $|x_1+x_2|$ the maximum of $|x_1+x_2|$,$|y_1+y_2|$ then we find that $|y_1+y_2| \leq |x_1+x_2|\leq |x_1|+|x_2|$ und thus $|x_1|+|x_2|$ is greater than the maximum of $|y_1+y_2|$ and $|x_1+x_2|$ and it follows
$max\{|y_1+y_2|,|x_1+x_2|\} \leq max\{|x_1|+|x_2|,|y_1|+|y_2|\}$
Let $m=max\{|x_1|,|x_2|\}$ and $m'=max\{|x_2|,|y_2|\}$ and then it is that $|x_1|+|x_2| \leq m + |x_2|$ because if $m=|x_1|$ then it is clear and if $m=|y_1|$ we have $|x_1| \leq |y_1|$ and we conclude $|x_1|+|x_2| \leq |y_1|+|x_2|=m+|x_2|$. We can apply this principle three more times and we have that
it is $|x_1|+|x_2| \leq m + |x_2| \leq m+m'$ and we also have $|y_1|+|y_2| \leq m + |y_2|\leq m+m'$ or in other words we find that
$max\{|x_1|+|x_2|,|y_1|+|y_2|\} \leq max\{|x_1|,|y_1|\}+max\{|x_2|,|y_2|\}$
So we finally find
$||v+w|| = max \{ |y_1+y_2|,|x_1+x_2|\} \leq max\{|x_1|+|x_2|,|y_1|+|y_2|\} \leq max\{ |x_1|,|y_1|\}+max\{|x_2|,|y_2|\}=||v||+||w||$
Thus we are done the norm is really a norm
$\endgroup$ 71 Answer
$\begingroup$Sorry, but this is still too complicated and verbose. I don't mean it's wrong, but it would be a very hard time reading it.
Check your attempt with the following. Writing clearer proofs requires time and effort, but don't be afraid.
Let's define $\|(x,y)\|=\max\{|x|,|y|\}$. We want to show that this defines a norm on $\mathbb{R}^2$.
We note first that, given $(x,y)\in\mathbb{R}^2$, we have $|x|\le\|(x,y)\|$ and $|y|\le\|(x,y)\|$.
Step 0. We observe that $\|(x,y)\|\ge0$, because $|x|\ge0$ and $|y|\ge0$. (Note: did you observe it?)
Step 1. Next we'll show that $\|(x,y)\|=0$ if and only if $(x,y)=(0,0)$. One direction is obvious, because $\max\{|0|,|0|\}=0$. For the other direction, if $\max\{|x|,|y|\}=0$, then both $|x|\le0$ and $|y|=0$, so $|x|=|y|=0$ and finally $x=y=0$.
Step 2. We want now to prove that $\|\lambda(x,y)\|=|\lambda|\,\|(x,y)\|$. Since $\lambda(x,y)=(\lambda x,\lambda y)$, we have$$ \|\lambda(x,y)\|=\max\{|\lambda x|,|\lambda y|\} =\max\{|\lambda|\,|x|,|\lambda|\,|y|\} $$If $|x|\le|y|$, then $|\lambda|\,|x|\le|\lambda|\,|y|$ and similarly if $|x|\ge|y|$. Hence$$ \max\{|\lambda|\,|x|,|\lambda|\,|y|\}=|\lambda|\max\{|x|,|y|\}=|\lambda|\,\|(x,y)\| $$
Step 3. For the triangle inequality, consider $v=(x_1,y_1)$ and $w=(x_2,y_2)$. We want to prove that $\|v+w\|\le\|v\|+\|w\|$.
By the standard triangle inequality we have$$ |x_1+x_2|\le |x_1|+|x_2|\le \|v\|+\|w\| \qquad |y_1+y_2|\le |y_1|+|y_2|\le \|v\|+\|w\| $$because of the initial remark.
Therefore$$ \|v+w\|=\max\{|x_1+x_2|,|y_1+y_2|\}\le\|v\|+\|w\| $$
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