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