# Can any integral of this form be written as a sum of convolutions?

Does the second equality always hold?

$$I(x) \equiv \int dy F(y,x-y) = \sum_{i=1}^{N}\int dy f_i(y)g_i(x-y)$$

Motivation: The first integral is not obviously a convolution that I could calculate quickly using FFTs. The second is a sum of convolutions that I could calculate quickly using FFTs. I can think of a possible counter-example or two but I don't know if I might not be able to write them as a sum of convolutions if I were clever enough about it.

• Explain your notation; what is $F$, what is $f_i$, what is $g_i$? Are you asking if a general function $F(y,x-y)$ can be written on the RHS form? – Winther Feb 25 '16 at 17:39
• Yes, they are general functions. $F(y,x-y)$ means that $x$ will never appear in the function on its own, but always in the combination $y-x$. – dbrane Feb 25 '16 at 17:43
• An example of $F$ that I cannot obviously recast into the RHS is $F(y,x-y)=\frac{1}{y^2+(x-y)^2}$ – dbrane Feb 25 '16 at 17:45
• What you're effectively asking is whether any two-variable function $F(w,z)$ 'separates' into a sum of products $F(w,z)=\sum_i f(w)g(z)$. I believe the answer is no; one reasonable candidate for failure is $F(w,z)=\sqrt{w+z}$ (in your notation, $F(y, x-y)=\sqrt{x}$). This function clearly can't be any rational combination of similar functions (unlike $e^t$ or $\sin(t)$); consider evaluating at $w=1, z=4$ where not only $w$ and $z$ but also $\sqrt{w}$ and $\sqrt{z}$ are all integers, but $F(w,z)=\sqrt{5}$ is irrational. IIRC this was studied during the early 20th century. – Steven Stadnicki Feb 25 '16 at 17:47
• Ok, that is a very convincing proof by contradiction. Thanks! – dbrane Feb 25 '16 at 17:55