# Convolution vanishes on an interval

Fix a "test" function $f(x)=x\exp(-x^2)$, which is nonzero except $x=0$. Suppose that $g$ is a function with some necessary regularity. Consider the convolution. $$(f\ast g )(x)=\int_{-\infty}^{+\infty} f(y)g(x-y)dy.$$ Assume that $(f\ast g) (x)=0$ on an open interval $x\in (a,b)$. Does this imply that $g(x)$ takes a constant value almost everywhere on $\mathbb{R}$? (The converse is true due to the symmetry of $f(x)$.)

Yes. Modifying things slightly to make some details come out nice:

The function $e^{-x^2/2}$ is its own Fourier transform. The transform of $xe^{-x^2/2}$ is $\pm i$ times the derivative of the transform of that gaussian, which is again essentially the function itself. Haven't got the $i$'s straight, so we take the absolute value: If $f(x)=xe^{-x^2/2}$ then $$|\hat f(\xi)|=|\xi|e^{-\xi^2/2}.$$

Now say $g\in L^1$. Then $$\widehat{f*g}=\hat f\hat g.$$And the function $\hat f\hat g$ tends to $0$ fast enough at infinity that its inverse transform extends to an entire function in the plane. An entire function that vanishes on an interval is constant.

To be specific, for every $y\in\Bbb R$ we have $$\int|\hat f(\xi)\hat g(\xi)|e^{y\xi}<\infty.$$

EDIT: The result holds under much weaker conditions on $g$. In particular it holds under conditions weak enough that $g$ doesn't have a Fourier transform. A way to see that the convolution extends to an entire function without using the Fourier transform is to simply define $$F(z)=\int g(t)(z-t)e^{-(z-t)^2/2};$$to a first approximation, any growth condition on $g$ that ensures that that integral converges for all $z\in\Bbb C$ should also suffice to make $F$ holomorphic. So for example if $\alpha>2$ then $$\int|g(t)|e^{-t^2/\alpha}<\infty$$should suffice.

But the condition $\int|fg|<\infty$ is not quite enough. This condition is not even enough to show that the convolution $f*g$ is defined on the line. The reason is that translates of $f$ are not comparable to $f$; note for example that $$e^{-(t-1)^2/2}=ce^te^{-t^2/2}.$$

• Can't $g$ be any function whose distributional Fourier transform is supported at $0$? The zero function is the only such function in $L^1$. – Justthisguy Aug 12 '15 at 1:45
• @AbrahamFrei-Pearson Certainly. I didn't mean to be giving the only conditions that would work, just one example of the OP's "some necessary regularity"... – David C. Ullrich Aug 12 '15 at 1:48
• @DavidC.Ullrich Thanks for your detailed answer. Would it be possible to extend the class of $g$ to those functions satisfying $\int |g| |f|dx<\infty$? – Uchiha Aug 12 '15 at 3:25
• @Ray No, but close. Edited answer. – David C. Ullrich Aug 12 '15 at 13:50
• @DavidC.Ullrich Thank you again. But could I ask one more question: how does $F(z)$ vanishing imply that $g$ is constant in the extended case? – Uchiha Aug 12 '15 at 18:40