Is there a measurable function $ f:\mathbb{R}\to \mathbb{R}^+ $ so that $ f\*f(x)=1 $ for all $ x\in \mathbb{R} $, i.e $$\int\limits\_{-\infty}^{\infty} f(t)f(x-t) dt=1 $$ for all $ x\in \mathbb{R} $.



Square root of dirac delta function
So I guess my initial intuition was wrong; there is enough "room at infinity" to concoct such a function $f$. The key lemma is



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> **Lemma.** Let $m\_1,m\_2,m\_3,\dots$ be an enumeration of the integers. Then there exists an increasing sequence $0 = f\_0 \leq f\_1 \leq f\_2 \leq \dots$ of finitely supported functions $f\_n: {\bf Z} \to {\bf R}^+$ such that for any $n \geq 0$, the function $f\_n\*f\_n$ equals one on $m\_1,\dots,m\_n$, and is bounded above by $1-2^{-n}$ for all other integers.
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Indeed, with such a sequence $f\_n$, if one sets $F := \sup\_n f\_n$, then by monotone convergence $F: {\bf Z} \to {\bf R}^+$ is such that $F\*F = 1$ on the integers, and if one then sets $f: {\bf R} \to {\bf R}^+$ to be $f(x) := F(\lfloor x \rfloor)$ (thus, $f$ is the convolution of $\sum\_n F(n) \delta\_n$ with $1\_{[0,1]}$, with $\delta\_n$ the Dirac delta at $n$) then $f\*f=1$ on the reals also.


One proves the proposition by a recursive construction on $n$. Clearly $f\_0=0$ obeys the required properties. Now suppose that $n \geq 1$ and that $f\_0,\dots,f\_{n-1}$ have already been constructed. Set $M$ to be a sufficiently large natural number (depending on $n$ and $f\_{n-1}$), and let $a\_1,\dots,a\_M$ be very large natural numbers that are *generic* in the sense that $(a\_1,\dots,a\_M)$ avoids a finite number of affine hyperplanes in ${\bf R}^M$. Set
$$ f\_n := f\_{n-1} + \sqrt{\frac{1-f\_{n-1}\*f\_{n-1}(m\_n)}{2M}} \sum\_{i=1}^M (\delta\_{a\_i} + \delta\_{m\_n - a\_i})$$
where $\delta$ now denotes the Kronecker delta.
Clearly $f\_n$ is finitely supported with $f\_n \geq f\_{n-1}$. If $M$ is large enough, and $a\_1,\dots,a\_M$ are in generic, one can verify that $f\_n\*f\_n - f\_{n-1} \* f\_{n-1}$ vanishes on $m\_1,\dots,m\_{n-1}$, equals $1 - f\_{n-1}\*f\_{n-1}(m\_n)$ on $m\_n$, and is bounded by $2^{-n}$ elsewhere, giving the required claim.


EDIT: It seems the basic reason why such a construction works is because convolution is not bounded on $\ell^\infty({\bf Z})$ (or on $L^\infty({\bf R})$); this unboundedness makes the problem significantly more underdetermined, and thus easier to solve. For instance, since convolution is bounded from $L^2({\bf R}) \times L^2({\bf R})$ to $L^\infty({\bf R})$, there is no solution to $f\*f=1$ with $f \in L^2({\bf R})$; indeed, a dense subclass argument or the Riemann Lebesgue lemma plus Plancherel shows that $f\*f \in C\_0({\bf R})$ whenever $f \in L^2({\bf R})$. In constrast, the function $f$ constructed above can be seen to lie in $L^{2,\infty}({\bf R})$ but no better, and convolution is not bounded from $L^{2,\infty}({\bf R}) \times L^{2,\infty}({\bf R})$ to $L^\infty({\bf R})$.


