# Square root of dirac delta function

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 $x\in \mathbb{R}$.

• euclid.ucc.ie/pages/staff/hanzon/Report2.pdf (in particular chapter 5) – Carlo Beenakker Apr 10 '16 at 15:37
• "Measurable" is not enough for convolution to exist. You should describe precisely what is your class of functions. – Alexandre Eremenko Apr 10 '16 at 15:46
• Yeah. But in this case its a positive function. So the integrations make sense. – DLN Apr 10 '16 at 15:48
• @Carlo Beenakker, in chapter 5, it has been proved that there is no real valued function whose square is the dirac delta function. But I can't exactly see how that solves this problem. I will read the chapter thoroughly and see if I can find something that helps. Thanks for the reference. – DLN Apr 10 '16 at 16:00
• Nice question! There is an a priori bound: for any interval $I$, one has $(\int_I f)^2 \leq \int_{I+I} f*f = |I+I| = 2|I|$, so $\int_I f \leq \sqrt{2} |I|^{1/2}$. In particular $\hat f$ exists as a tempered distribution at least. Unfortunately there is not yet enough regularity to justify the equation $\hat f^2 = \delta$. Nevertheless the answer should still be negative, I think. – Terry Tao Apr 10 '16 at 16:40

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

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.

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 $L^\infty({\bf R})$.

• Another remark is that once one has an $f$, then $g=\varphi*f$ for a non-negative $\varphi\in C_0^{\infty}$ with $\int\varphi =1$ is a smooth function with $g*g=1$. – Christian Remling Apr 10 '16 at 17:29
• I'm now curious about what $\widehat{f}\in\mathcal S'$ looks like, I wonder if anything meaningful can be said. – Christian Remling Apr 10 '16 at 18:36
• I think it can be placed in the Besov space $B^{0,\infty}_2$ (a Fourier-analytic version of "weak $L^2$"). Almost, but not quite, enough regularity to define $(\hat f)^2$ in any reasonable sense. – Terry Tao Apr 10 '16 at 19:03