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I was working on this problem, and I thought I'd post my answer so people could see if they have a better one:

Spivak Calculus, 4th ed., problem 3-18:

Suppose $f,\,g,\,h,\,k$ are functions from $\mathbb{R}$ to $\mathbb{R}$. Precisely what conditions must $f,\,g,\,h,\,$ and $k$ satisfy in order that $$ f(x)g(y)=h(x)k(y) \tag{1} $$ for all $x,\,y\,\in \mathbb{R}$?

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Condition (1) is equivalent to saying that the matrix $$A(x,y)=\begin{pmatrix} f(x) & h(x) \\ k(y) & g(y) \end{pmatrix}$$ is singular for all $x,y$. This is equivalent to its columns being linearly dependent: i.e., for all $x,y$ there exist $a,b$ (not both zero) such that $$af(x)+bh(x) =0 = ak(y)+bg(y)$$ In principle, $a $ and $b$ could depend on $x$ and $y$. To clarify the matter, consider two cases.

  1. One of the rows of $A$ is identically zero. Then there are no conditions on the other row; property holds.
  2. Both rows are sometimes nonzero. Then we can choose $a,b$ above independent of $x,y$. Indeed, fix $x,y$ such that $A(x,y)$ has nonzero rows, and find $a,b$ for this matrix. The same $a,b$ must work for every other $A(x',y' )$, as we can see by considering $A(x',y)$ and $A(x,y')$ first.

Summary: (1) holds if and only if one of the following is true:

  1. $f\equiv 0 \equiv h$;
  2. $g\equiv 0 \equiv k$;
  3. There exist constants $a,b$, not both zero, such that $af+bh\equiv 0\equiv ak+bg$.
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  • $\begingroup$ Very good answer. Thanks as always $\endgroup$ – Eric Auld Apr 21 '14 at 2:43
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First of all there is the possibility that both sides of (1) are identically zero. For this it is necessary that $$ \text{One of $f,\,g$ and one of $h,\, k$ are identically zero} $$ since if not, we can find $x,y$ such that one side is nonzero.

If neither side is identically zero, then let $Z_f=\{x \in \mathbb{R}\mid f(x)=0\}$, and similarly $Z_g,\,Z_h,\,Z_k$.

Observe: If neither side of (1) is identically zero, we have $$ Z_f = Z_h \quad \text{ and } \quad Z_g=Z_k, $$ and $$ \frac{f}{h} \text{ is constant on $(Z_f)^c$ and } \frac{k}{g} \text{ is constant on $(Z_k)^c$, and they are the same constant.} $$

This seems to be the most we can say. Anyone see anything else?

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  • $\begingroup$ The constant should be the same for both ratios. And then you have "if and only if" condition, so there's nothing else to say. $\endgroup$ – user127096 Apr 20 '14 at 15:40
  • $\begingroup$ @HowAboutaNiceBigCupof I agree the constant should be the same...thank you! I'm not clear on what you mean by "if and only if". Can you expand on this? $\endgroup$ – Eric Auld Apr 20 '14 at 15:44

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