0
$\begingroup$

My question is to

determine the region of $\Bbb R^2$ on which $f(x,y)=\max(x,y)$ is totally differentiable.

So far I have proved that $\max(x,y)$ is continuous on $\Bbb R^2$ because $\max(x,y)=\frac{x+y}2+\frac{|x-y|}2$ is the sum of continuous functions on $\Bbb R^2$. I guess $\max(x,y)$ is totally differentiable $\iff$ $x \ne y$ but I got stuck at the very first step. I wanted to show that if $x \ne y$ then $\max(x,y)$ is totally differentiable. Since $x \ne y$, $$f(x,y)=\begin{cases}x & \text{if} \ x>y\\ y & \text{if} \ x<y. \end{cases}$$ Symbolically, $$\frac{\partial f}{\partial x} = \begin{cases} 1, & x>y\\ 0, & x<y. \end{cases}$$

$$\frac{\partial f}{\partial y} = \begin{cases} 0, & x>y\\ 1, & x<y. \end{cases}$$ But for some fixed $(a,b)\in \Bbb R^2$ with $a>b$, I cannot convince mysefl why $$\frac{\partial f}{\partial x}(a,b)=\lim_{t \to 0}\frac{\max(a+t,b)-a}t=1?$$ Also $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ seem not to be continuous. Nothing guarantees the differentiability of $f(x,y)$ here.

Any help would be much appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ Use the fact that if $a > b$, then $a + t > b$ for sufficiently small $|t|$. $\endgroup$ – anomaly Oct 4 '16 at 16:33
  • $\begingroup$ I see, now I understand that $f_x,f_y$ exist when $x\ne y$. $\endgroup$ – user Oct 4 '16 at 16:47
3
$\begingroup$

In the open set $U = \{(x,y): y>x\},$ $f(x,y) = y.$ Of course the function $y$ is totally differentiable everywhere. Therefore $f$ is totally differentiable on $U.$ (No need to compute any partial derivatives.) The same holds for $\{(x,y): y<x\}.$

We're left thinking about the set $E= \{(x,y): y=x\}.$ Let $(a,a) \in E.$ If $h>0,$ we have

$$\frac{f(a,a+h)-f(a,a)}{h} = \frac{a+h - a}{h} = 1.$$

If $h<0,$ we have

$$\frac{f(a,a+h)-f(a,a)}{h} = \frac{a - a}{h} = 0.$$

It follows that $\lim_{h\to 0}(f(a,a+h)-f(a,a))/h$ does not exist, i.e., $\partial f/\partial y (a,a)$ does not exist. Thus $f$ is not differentiable at any point of $E.$

$\endgroup$
1
  • $\begingroup$ I just want to get the full understanding by trying to prove that $f(x,y)=y$ is differentiable everywhere on your open set $U$. Fix $(a,b)\in \Bbb R^2$ with $a>b$, we can compute $\frac{\partial f}{\partial x}(a,b)=0, \frac{\partial f}{\partial y}(a,b)=1$ and $\lim_{(x,y) \to (a,b)}\frac{\max(x,y)-y}{\sqrt{(x-a)^2+(y-b)^2}}=0$. Thus $f(x,y)=y$ is differentiable everywhere on $U$. Is that correct? $\endgroup$ – user Oct 5 '16 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.