9
$\begingroup$

I am working through Ted Shifrin's book Multivariable Mathematics. There is an exercise problem that is meant to demonstrate that one can have $\dfrac{\partial^2 f}{\partial x \partial y} = 0$ but$ f(x,y) \neq g(x) + h(y)$.

The question (3.6.11) is as follows: $$ \mathrm{Given} \; f(x, y) = \begin{cases} 0, \; x < 0\; \mathrm{or} \; y < 0 \\x^3, \; x \geq 0 \; \mathrm{and} \; y > 0 \end{cases} $$

  1. Show that $f$ is $C^2$

  2. Show that $\dfrac{\partial^2 f}{\partial x \partial y} = 0$

  3. Show that $f(x,y)$ cannot be written as $g(x) + h(y)$ for appropriate functions $g, h$.


I see that the domain is the entire plane except the x-axis, that is $\mathbb{R}^2 -\{y=0\}$. The function is then 0 in all quadrants except the first, where it is $x^3$.

I could show 1. and 2. above, but I am puzzled by two things.

Q1 What's wrong with writing $f(x,y) = g(x) + h(y)$ piecewise in each quadrant ?

Q2. Will anything change if the domain allows the line $y=0$ also ?

Q3. What is the takeaway from this problem ? I do not understand that.

$\endgroup$
  • $\begingroup$ For Q2, note that $lim_{y\rightarrow0^{+}}f(1,y)=1$, whereas $lim_{y\rightarrow0^{-}}f(1,y)=0$, so the function cannot be continuously extended to the positive x-axis. The function is already defined equal to 0 on the negative x-axis, by the first piecewise condition. $\endgroup$ – Leland Reardon Nov 16 '18 at 21:57
3
$\begingroup$

For Q1; defining the quadrants requires both $x$ and $y$. For example, defining $$g(x):=\left\{\begin{array}{ll} 0 &\text{ if } x<0\ \text{ or } \; y < 0, \\ x^3 &\text{ if } x\geq0 \; \text{ and } \; y > 0. \end{array}\right.,$$ does not make sense as now $g(x)$ depends on $y$. More formally; if $f(x,y)=g(x)+h(y)$ then for all $(x_1,y_1),(x_2,y_2)\in\Bbb{R}^2$ we have $$f(x_1,y_1)+f(x_2,y_2)-f(x_1,y_2)-f(x_2,y_2)=0.$$ But this is clearly not the case, take for example $x_1=y_1=1$ and $x_2=y_2=-1$.


For Q2; the given function $f(x,y)$ does not extend to a function differentiable on $\Bbb{R}^2$. There are no functions with this property that are differentiable on all of $\Bbb{R}^2$. See also this excellent answer.


For Q3; the takeaway is that the implication $$\frac{\partial^2f}{\partial x\partial y}=0 \qquad\Rightarrow\qquad f(x,y)=g(x)+h(y),$$ depends crucially on the domain; it holds if the domain is $\Bbb{R}^2$, but not if the domain is, for example, a disconnected subset of $\Bbb{R}^2$.

$\endgroup$
  • $\begingroup$ Could you provide an example of a function like that defined on all of $\mathbb{R}^2$ ? $\endgroup$ – me10240 Nov 16 '18 at 22:51
  • $\begingroup$ I'm sorry, I was mistaken. It must be time for bed, let me correct my answer first. $\endgroup$ – Inactive - avoiding CoC Nov 16 '18 at 23:06
  • $\begingroup$ The bit about the domain of g(x) needing to be described by y is just mindblowing, many thanks to you for the answer, and to Professor Shifrin for the question. Its easy to forget that a function and its domain are inseparable. $\endgroup$ – me10240 Nov 17 '18 at 1:56
  • $\begingroup$ I think Q1 has to be answered in affirmative. E.g. for the first quadrant $f(x,y)=x^3+0$, so is $g(x)=x^3$ and $h(y)=0$ $\endgroup$ – Rafa Budría Nov 17 '18 at 11:13
  • $\begingroup$ @RafaBudría I do not understand; Q1 is not a yes/no question. $\endgroup$ – Inactive - avoiding CoC Nov 17 '18 at 11:22

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.