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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.

For Q1; defining the quadrants requires both $x$ and $y$. For example, defining
$$g(x):=left{begin{array}{ll}
0 &text{ if } x<0 text{ or } ; y < 0, \
x^3 &text{ if } xgeq0 ; text{ and } ; y > 0.
end{array}right.,$$
does not make sense as now $g(x)$ depends on $y$. More formally; if $f(x,y)=g(x)+h(y)$ then for all $(x_1,y_1),(x_2,y_2)inBbb{R}^2$ 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
$$frac{partial^2f}{partial xpartial y}=0
qquadRightarrowqquad
f(x,y)=g(x)+h(y),$$
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 $Bbb{R}^2$.