# Does the Banach algebra $L^1(\mathbb{R})$ have zero divisors?

Assume that the functions $f,g: \mathbb{R}\rightarrow \mathbb{R}$ are integrable and equal to zero on $(-\infty,0)$, (i.e $f,g \in L^+$). Then by Titchmarsh's theorem:

$f*g$ is zero almost everywhere iff $f$ or $g$ is zero almost everywhere.

Hence the Banach subalgebra $L^+$ of $L^1(\mathbb{R})$ has no zero divisors.

Is the same true for arbitrary integrable $f$ and $g$ on $\mathbb{R}$ ?

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 See also this related MO thread – t.b. Dec 11 '11 at 23:04

To make Yemon Choi's idea more explicit:

The bump function

$$b(x):=\cases{\exp{-1\over 1-x^2} & \bigl(|x|<1\bigr) \cr 0 & \bigl(|x|\geq 1\bigr)\cr}$$

is $C^\infty$ and has compact support $[-1,1]$; whence it is in the Schwartz space ${\cal S}$ and is the inverse transform of its Fourier transform $\hat b\in {\cal S}$. The same is true for the functions $f_1(x):=b(x-1)$ and $f_2(x):=b(x+1)$ whose product is identically zero. But up to a constant factor this product is equal to $\hat f_1\ *\ \hat f_2$.

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 Thanks a lot for answer. – L.T Dec 13 '11 at 13:22

The MO thread that t.b. links to doesn't seem to quite answer the original question, which if I've understood it correctly is:

do there exist $f,g\in L^1({\mathbb R})$, neither of which is a.e. zero, such that $f*g$ is a.e. zero?

I am pretty sure that the answer is yes, because the answer to the analogous question for $\ell^1({\mathbb Z})$ is yes. To see this, work on the circle and consider the simplest function $h_1$ which is piecewise linear on $[-\pi,\pi]$, is zero on the intervals $[-\pi,-\delta]$ and $[\delta,\pi]$ for some fixed $\delta$, and which is $1$ at $0$. (Triangular spike.) Then viewed as a continuous function on ${\mathbb R}/2\pi{\mathbb Z}={\mathbb T}$, the Fourier series of $h_1$ is absolutely summable. Now let $h_2(x) = h_1(x+\pi)$, clearly $h_1\cdot h_2=0$, which means that (FS of $h_1$) $*$ (FS of $h_2$) $=0$.

Now your question takes place on the line, not the circle, but my feeling is that the same construction should work. That is, let $h_1$ and $h_2$ be the same functions as before, but regard them as functions on ${\mathbb R}$. Then both should be the Fourier transforms of integrable functions $f_1,f_2\in L^1{\mathbb R})$, and once again $f_1*f_2=0$ since $h_1\cdot h_2=0$.

(Actually, thinking about it a bit more, I "know" that the answer to your question must be yes, because it is equivalent to the question "does the Fourier algebra $A({\mathbb R})$ contain non-trivial zero divisors?", and it's known that $A({\mathbb R})$ is a regular Banach algebra of functions, meaning that one can find functions which are zero on any prescribed closed subset $F$ and 1 on any prescribed compact set disjoint from $F$; pointwise product in $A({\mathbb R})$ corresponds to convolution in $L^1({\mathbb R})$.)

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 Thanks it is helpful. – L.T Dec 12 '11 at 7:58