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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 16, 2018 at 21:57
  • $\begingroup$ The domain is the entire plane minus the nonnegative part of the $x$-axis. The negative part of the $x$-axis is included in the first condition “$x<0$ or [...]”, so $f$ is defined (equal to zero) there. $\endgroup$ Oct 7, 2021 at 10:36

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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 $\Bbb{R}^2$.

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  • $\begingroup$ Could you provide an example of a function like that defined on all of $\mathbb{R}^2$ ? $\endgroup$
    – me10240
    Nov 16, 2018 at 22:51
  • $\begingroup$ 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, 2018 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 $h(y)=0$ $\endgroup$ Nov 17, 2018 at 11:13
  • $\begingroup$ @RafaBudría I do not understand; Q1 is not a yes/no question. $\endgroup$ Nov 17, 2018 at 11:22
  • $\begingroup$ Answering in affirmative to the question Q1 is, to my understanding, say that it is correct writing $f(x,y) = g(x) + h(y)$ piece wise in each quadrant. $\endgroup$ Nov 17, 2018 at 11:30

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