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Let $S$ be a non-empty open subset of $\mathbb{R}^2$, let $B$ be a set (possibly empty) of boundary points of $S$, and let $D = S\cup B$.

Is it possible to have single-variable functions $f_1:\mathbb{R}\longrightarrow\mathbb{R}$ and $f_2:\mathbb{R}\longrightarrow\mathbb{R}$ such that $$f_1(y)-f_2(x)=(x+x^3)y, \text{ for all} \ (x,y)\in D \ ?$$

3 Answers 3

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No, it's not possible.

Suppose that $f_1$ and $f_2$ exist. Let $(x_1,\overline y)$ and $(x_2,\overline y)$ be two points in $S$ such that $ b_1 =x_1+x_1^3\ne x_2+x_2^3=b_2$, and call $a_1=f_1(x_1)$ and $a_2 = f_1(x_2)$. We have that $$ f_2(y) = -yb_1+a_1\qquad f_2(y) = -yb_2+a_2 $$ in an open neighborhood of $\overline y$, but the two functions are different, since $b_1\ne b_2$.

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Suppose it were possible.

Then $f_1(0) - f_2(x) = (x + x^3)*0 = 0$ and $f_2(x) = -f_1(0)= c$ for all $x$. So $f_2(x)$ is a constant function.

Then $f_1(y) - c = (x+x^3)*y$ and $f_1(y) = (x+x^3)*y+c$ is not well defined as it will have different values for different $x$ which are not variables of the function.

... or ... $f_1(y) - f_2(0) = 0*y = 0$ so $f_1(y)=f_2(0) = c$ is also a constant function.

So $f_1(y) - f_2(x) = -c + c = 0$ is a constant function. Which $(x+x^3)y$ clearly is not.

So, no. It is not possible.

....

Or to put it more simply:

$f_1(0) - f_2(0) = 0$

$f_1(0) - f_2(1) = 0$

$f_1(1) - f_2(0) = 0$

$f_1(1) - f_2(1) = 1$.

So we must solve four equations with four unknowns:

$a - b = 0$

$a - d = 0$

$c - b = 0$

$c - d = 1$.

$a = b; a = d; c=b=a=d; c-d = c- c= 0 = 1$. Impossible.

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    But, we do not know if the points of the type $(0,y)$, or, $(x,0)$ belong to the set $D$! – 2017-01-09
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    Good point! I didn't take that into account at all. But, well... If $(a,b) \in S$ and $N$ is an open neighborhood subset of D then we can find $\epsilon$ and $\delta$ so that $((x+\epsilon)+(x + \epsilon)^3)(y-\delta) = (x + x^3)y$ and $(x+\epsilon,y-\delta) \in N$. We can work out a similar contradiction. I think. It would take a bit of work though. Well.... Exod's answer works. – 2017-01-09
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If we assume $f_1$ and $f_2$ are differentiable, and take the derivative with respect to $y$ of your last equation, we get $f_1^{\prime}(y) = x+x^3.$ So a function of $y$ alone somehow equals a function of $x$ alone.

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    Why would we assume $f_1$ and $f_2$ are differentiable? – 2017-01-09
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    @fleablood Because sometimes, the way to solve a problem is to first solve a more specialized one. It gives insight into the relationships in the problem and leads us to the more general solution. Is there some other way to do math? – 2017-01-09