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$\renewcommand\C{\mathbb C}\newcommand{\R}{\mathbb R}$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.

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 $x=(x_1,\dots,x_n)\in\R^n$.

$\renewcommand\C{\mathbb C}\newcommand{\R}{\mathbb R}$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.)

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 $x=(x_1,\dots,x_n)\in\R^n$.

$\newcommand\C{\mathbb C}$Schwartz's Paley–Wiener theorem states the following:

An entire function $F$ on $\C^n$ is the Fourier–Laplace transform of a distribution $v$ of compact support if and only if for all $z\in\C^n$ $$|F(z)| \le C (1 + |z|)^N e^{B|\Im z|} \tag{1}\label{1}$$ for some real constants $C>0,N,B$. The distribution $v$ in fact will be supported in the closed ball of center $0$ and radius $B$.

In our case, $v:=u*u$ is compactly supported. So, by Schwartz's Paley–Wiener theorem, \eqref{1} with the Fourier–Laplace transform $\hat v=\hat u^2$ of $v$ in place of $F$ holds for some real constants $C>0,N,B$ and all $z\in\C^n$. Therefore and because $|\hat u|=\sqrt{|\hat v|}$, \eqref{1} with the Fourier–Laplace transform $\hat u$ of $u$ in place of $F$ holds with constants $\sqrt C,N/2,B/2$ in place of $C,N,B$ and all $z\in\C^n$. So, $u$ will be supported in the closed ball of center $0$ and radius $B/2$.

$\renewcommand\C{\mathbb C}\newcommand{\R}{\mathbb R}$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.

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 $x=(x_1,\dots,x_n)\in\R^n$.

$\newcommand\C{\mathbb C}$Schwartz's Paley–Wiener theorem states the following:

An entire function $F$ on $\C^n$ is the Fourier–Laplace transform of a distribution $v$ of compact support if and only if for all $z\in\C^n$ $$|F(z)| \le C (1 + |z|)^N e^{B|\Im z|} \tag{1}\label{1}$$ for some real constants $C>0,N,B$. The distribution $v$ in fact will be supported in the closed ball of center $0$ and radius $B$.

In our case, $v:=u*u$ is compactly supported. So, by Schwartz's Paley–Wiener theorem, \eqref{1} with the Fourier–Laplace transform $\hat v=\hat u^2$ of $v$ in place of $F$ holds for some real constants $C>0,N,B$ and all $z\in\C^n$. So, \eqref{1} with the Fourier–Laplace transform $\hat u$ of $u$ in place of $F$ holds with constants $\sqrt C,N/2,B/2$ in place of $C,N,B$ and all $z\in\C^n$. So, $u$ will be supported in the closed ball of center $0$ and radius $B/2$.

$\newcommand\C{\mathbb C}$Schwartz's Paley–Wiener theorem states the following:

An entire function $F$ on $\C^n$ is the Fourier–Laplace transform of a distribution $v$ of compact support if and only if for all $z\in\C^n$ $$|F(z)| \le C (1 + |z|)^N e^{B|\Im z|} \tag{1}\label{1}$$ for some real constants $C>0,N,B$. The distribution $v$ in fact will be supported in the closed ball of center $0$ and radius $B$.

In our case, $v:=u*u$ is compactly supported. So, by Schwartz's Paley–Wiener theorem, \eqref{1} with the Fourier–Laplace transform $\hat v=\hat u^2$ of $v$ in place of $F$ holds for some real constants $C>0,N,B$ and all $z\in\C^n$. Therefore and because $|\hat u|=\sqrt{|\hat v|}$, \eqref{1} with the Fourier–Laplace transform $\hat u$ of $u$ in place of $F$ holds with constants $\sqrt C,N/2,B/2$ in place of $C,N,B$ and all $z\in\C^n$. So, $u$ will be supported in the closed ball of center $0$ and radius $B/2$.