Level sets for the function $f$ Describe the graph of the function $$f: \mathbb{R}^2 \to \mathbb{R}, (x, y) \to \max (|x|, |y|)$$ computing some level sets and some intersections.
I have done the following:
The level curves are defined by $$\{(x, y) \mid \max (|x|, |y|)\}.$$

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*For $c=0$ we have tha $\max (|x|, |y|)=0 \Rightarrow x=y=0$. So, for $c=0$ the level set consists of the axis $y$ and $x$.


*For $c<0$, the level set is the empty set.
Is this correct?
Could I improve something?

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*For $c>0$ we have that $\max (|x|, |y|)=c>0$. How can we find of what the level set consists in this case? Do we have to suppose that for example $|x|$ is the maximum?

 A: Its graph looks like a pyramid-shaped sink:

Try using also Wolfram Alpha. A general suggestion for the analysis of one- and two-variable functions: Once you get a geometric idea, so that you can literally see what you're supposed to be proving, the rigorous analytics will become way simpler.

As for the analytics: you already concluded that the level set
$$\mathcal L_c\equiv\{(x,y)\in\mathbb R^2\,|\,\max\{|x|,|y|\}=c\}$$
is empty if $c<0$ and is the singleton $\{(0,0)\}$ if $c=0$. Now if $c>0$, then $\max\{|x|,|y|\}=c$ if and only if


*

*$|x|\geq |y|$ and $|x|=c$; or

*$|x|\leq|y|$ and $|y|=c$.


The first case holds if and only if (a) $x=c$ and $y\in[-c,c]$; or (b) $x=-c$ and $y\in[-c,c]$. The second case holds if and only if (c) $y=c$ and $x\in[-c,c]$; or $y=-c$ and $x\in[-c,c]$. In fact, each of the cases (a) to (d) describes one edge of a square in the plane. The four vertices of this square are $(c,c)$, $(c,-c)$, $(-c,c)$, and $(-c,-c)$. Therefore, the level set $\mathcal L_c$ is the boundary (perimeter) of this square.
For example, if you were to drag your finger along the edge of the sink in the picture above (carefully, it may be sharp), you would just be tracing out the level set corresponding to the height of the sink.
A: First, if $c<0$ the level set is obviously empty, since $|x|\ge 0$ and $|y|\ge 0$.
Second, if $c=0$, then the level set is $\{(x,y):\max(|x|,|y|)=0\}=\{(0,0)\}$, i.e. just the origin.
Last, if $c>0$, then we will need to consider several cases (in fact, they are all the same thanks to the symmetry of the problem).
Suppose that $x\ge |y|\ge 0$, then $ f(x,y)=x$ leads to the level set $\{(x,y):0\le |y|\le x\le c\}$, i.e. a triangle with vertices $(0,0)$, $(c,c)$, $(c,-c)$.
By conducting the similar analysis for remaining cases ($0\le |x|\le y$, $x\le-|y|\le 0$, and $y\le -|x|\le 0$) you will obtain that the level set in total is a square with vertices $(\pm c,\pm c)$.
