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Consider a function of the form $g(x,y) = f(x) f(y)$. This function can just as well be written as $h(x-y) k(x+y)$ for some functions $h$ and $k$. Does the form of this function put any restriction on $h(x-y)$, or can $h(x-y)$ be an arbitrary function given a suitable choice of $f$?

Henry Shackleton
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1 Answers1

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You cannot always decompose $g(x,y)$ using your suggestion of it being a product of $2$ alternate functions. Consider the quite simple case of $f(x) = x$. You get

$$g(x,y) = f(x)f(y) = xy \tag{1}\label{eq1A}$$

Consider functions $h(x-y)$ and $k(x+y)$ existing such that

$$g(x,y) = h(x-y)k(x+y) = xy \tag{2}\label{eq2A}$$

Set $y = x$ to get

$$h(0)k(2x) = x^2 \implies h(0)k(x) = \left(\frac{x}{2}\right)^2 \implies k(x) = \frac{x^2}{4h(0)} \tag{3}\label{eq3A}$$

Substituting this into \eqref{eq2A} gives

$$h(x-y) = \frac{xy}{k(x+y)} = \frac{4h(0)xy}{(x+y)^2} \tag{4}\label{eq4A}$$

Using $x = 2y$ gives

$$h(y) = \frac{4h(0)(2y^2)}{9y^2} = \left(\frac{8}{9}\right)h(0) \implies h(0) = \left(\frac{8}{9}\right)h(0) \implies h(0) = 0 \tag{5}\label{eq5A}$$

However, from \eqref{eq3A}, this gives that $x^2 = 0$, which is incorrect.

This provides one simple example where you can't decompose a function differently as you request. I believe there are many other such cases (in fact, most of the time), but I'm unsure how to classify, in general, when it will, and will not, work.

In fact, I have trouble coming up with any non-trivial examples which do work. Nonetheless, one relatively simple one is $f(x) = e^x$, so $f(x)f(y) = \left(e^x\right)\left(e^y\right) = e^{x+y}$. You can then have $h(x-y) = 1$ and $k(x+y) = e^{x+y}$.

John Omielan
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  • I see, thank you for the explanation. My motivation for asking this question was to understand functions of the form $\int dx dy f(x) f(y) e^{i(x-y)t}$ as a function of $t$. I thought this could be done by recasting it as a function of $x-y$ and doing a Fourier transform, but it looks like this isn't generally possible. I'll try to phrase my question properly and resubmit. – Henry Shackleton Dec 20 '19 at 22:42
  • @HenryShackleton You're welcome. However, it seems what you're looking for is quite different from what you asked here, so instead of resubmitting this question, which is a fairly interesting one as it is, I suggest you create a new question and, if you wish & think it's appropriate, reference this question. – John Omielan Dec 20 '19 at 22:44
  • In the second equality in $(4)$, you use that $k(x+y)=\frac{x^2}{4h(0)}$. But what you established in $(3)$ was $k(x)=\frac{x^2}{4h(0)}$. – MathematicsStudent1122 Dec 20 '19 at 22:44
  • @MathematicsStudent1122 Thank you for pointing that out. You're correct. I had just noticed that & was fixing it when you made your comment. However, I've now fixed it, with the same general conclusion occurring. – John Omielan Dec 20 '19 at 22:49
  • Examining the cases x=y, And x=-y, and assuming without loss of generality that k(0)=1 allows you to establish formulas for h and k in terms of f, then you can say that such a decomposition is possible when f satisfies a resulting functional equation. – Robo300 Dec 20 '19 at 22:51
  • @JohnOmielan Yes, sorry, that was what I meant by "resubmit" - apologies for the confusion. – Henry Shackleton Dec 20 '19 at 22:53