Convolution of $L^2$ functions

Question

Let $u\in L^2(\mathbb R^n)$: then $u\ast u$ is a bounded continuous function. Let me assume now that $u\ast u$ is compactly supported. Is there anything relevant that could be said on the support of $u$, in particular in the case where $n\ge 2$?

Answer 1

$u$ need not be compactly supported.

Take any $\varphi\in C_0^{\infty}$, $\varphi\not\equiv 0$, and let $\widehat{u}=\widehat{\varphi}^{1/2}$ be any non-holomorphic (but measurable) square root. (Typically, $\widehat{u}$ will fail to be holomorphic automatically, when $\widehat{\varphi}$ has zeros.) Clearly $\widehat{u}\in L^2$, since $\widehat{\varphi}$ is a Schwartz function.

Then $(u*u)\widehat{\:\:\:}=\widehat{u}\,^2=\widehat{\varphi}$, so $u*u=\varphi$ is compactly supported, but $u$ isn't because its Fourier transform is not entire.

We can also produce explicit examples in this fashion, if we start out with a (non-smooth) $\varphi$ whose Fourier transform we can compute. For example, $u(x)=\log |(x-1)/(x+1)|$ works; this comes from a triangular $\varphi$.

Answer 2

In one dimension, observe from inspection of the Fourier symbol of the Hilbert transform $H$ that one has the identity $Hu * Hu = - u*u$ for any $u \in L^2$. So one can easily generate a non-compactly supported $u \in L^2({\bf R})$ for which $u*u$ is compactly supported by starting with any bump function of non-zero mean and applying the Hilbert transform to get a function that decays like $1/x$ at infinity.

More generally, in higher dimensions we have $T_m u * T_m u = u * u$ whenever $T_m$ is a Fourier multiplier whose symbol $m$ takes values in $\{-1,+1\}$. By taking a highly discontinuous choice for $m$ and applying $T_m$ to a bump function one can create $u \in L^2$ that decay extremely slowly at infinity (in fact I think any decay consistent with square-integrability could be permitted, though I have not checked this carefully) but for which $u*u$ remains compactly supported (and even smooth).

Answer 3

$\renewcommand\C{\mathbb C}\newcommand{\R}{\mathbb R}$To complement the answer by Christian Remling, let us provide explicit examples of $u\in L^2(\R^n)$ such that $u*u$ is compactly supported but $u$ is not compactly supported.

Consider first the case $n=1$. Let $w(x):=\frac12\,1(|x|<1)$ for real $x$. Then $w\in L^2(\R)$ and for the Fourier transform $\hat w$ of $w$ and all real $t\ne0$ we have \begin{equation*} \hat w(t)=\int_\R dx\, e^{itx} w(x)=\frac{\sin t}t, \tag{1}\label{1} \end{equation*} with $\hat w(0)=1$. Let now $v:=w*w$. Then $v$ is of course compactly supported, and \begin{equation*} \hat v=\hat w^2. \end{equation*} Next, let $u$ be the inverse Fourier transform of $|\hat w|=\sqrt{\hat v}$, so that $u$ is the limit in $L^2(\R)$ as $A\to\infty$ of the functions $u_A$ defined by the formula \begin{equation*} u_A(x):=\frac1{2\pi}\,\int_{-A}^A dx e^{-itx} \frac{|\sin t|}{|t|} =\frac1{2\pi}\,\int_{-A}^A dx \cos tx\, \frac{|\sin t|}{|t|} \end{equation*} for real $x$. Note that $u\in L^2(\R)$, since $\hat u=|\hat w|\in L^2(\R)$. (One may also note that the function $u$ is real-valued.)

Also, $u*u=v$, since $\hat u^2=|\hat w|^2=\hat v$. So, $u*u$ is compactly supported.

If the function $u$ were compactly supported, then, by the easy part of Schwartz's Paley–Wiener theorem, the Fourier transform $\hat u=|\hat w|$ could be extended to an entire function -- which is impossible, because, in view of \eqref{1}, the function $|\hat w|$ is not smooth on $\R$.

Thus, $w\in L^2(\R)$ and $u*u$ is compactly supported, but $u$ is not compactly supported. (In fact, $u$ is not even in $L^1(\R)$ -- because then $\hat u=|\hat w|$ would be differentiable, which it is clearly not.)

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To get now, for any natural $n$, an explicit example of a function $U\in L^2(\R^n)$ such that $U*U$ is compactly supported but $U$ is not compactly supported, it is enough to tensorize the function $u$ from the "one-dimensional" example above: \begin{equation*} U(x):=u^{\otimes n}(x)=u(x_1)\cdots u(x_n) \end{equation*} for $x=(x_1,\dots,x_n)\in\R^n$.