Does $L^2(\mathbb{R})$ have idempotent under convolution operation.

Question

One can easily prove that there does not exist any $f\in L^1(\mathbb{R})$ such that $f*f=f$ holds due to Riemann-Lebesgue Lemma. But what about the case when $f\in L^2(\mathbb{R})?$ Is there any counterexample or is there any such function which fails to exist?

Answer 1

Assume $f\in L^2(\Bbb R^d)$ is a solution to $f*f = f$. Then $g:=\mathcal F(f)\in L^2$ by Plancherel and in the sense of distributions (see the Lemma below), $$ g^2 = g $$ Since $g\in L^2$, one deduces that the above identity holds a.e. and so $g$ is an indicator function. Since $g\in L^2$, it is the indicator function of a set of finite measure.

Conversely, if $g$ is the indicator function of a set of finite measure, then $g\in L^1$ and $g^2\in L^1$ and $g^2=g$, so taking the Fourier transform in the classical sense, $f*f=f$.

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Lemma: if $f\in L^2$, then in the sense of distributions, $\mathcal F(f*f) = \mathcal F(f)^2$.

Proof: let $f_n$ be a sequence of $C^\infty_c$ functions converging to $f$ in $L^2$. Then $\mathcal F(f_n)^2\to \mathcal F(f)^2$ in $L^1$ (because $a^2-b^2=(a-b)(a+b)$ and by Cauchy-Schwarz), and so in the sense of (tempered) distributions.

Moreover, $f_n*f_n$ converges to $f*f$ in $L^\infty$ by Young's inequality (write $f_n*f_n-f*f = f_n*(f_n-f) + (f_n-f)*f$), therefore, for any Schwartz function $\varphi$, $\langle f_n*f_n,\mathcal F \varphi \rangle \to \langle f*f,\mathcal F \varphi \rangle$, that is $\mathcal F(f_n*f_n)\to \mathcal F(f*f)$ in the sense of distributions.

Finally, for every $n$, $\mathcal F(f_n)^2 = \mathcal F(f_n*f_n)$, so taking the limit in the sense of distributions, $\mathcal F(f)^2 = \mathcal F(f*f)$

Answer 2

So, I came up with a solution. Intuitively if I take the fourier transformation on both side of the equation \begin{align} f*f=f, \end{align} then it suggest that $\hat{f}$ is sort of indicator function of a set $A$ of finite measure. Now, I come to the rigorous solution, let's take $g=\chi_A$, where $\lambda(A)<\infty$. Now take the inverse Fourier transform $F^{-1}(g)$ and let's call it $f$. Clearly $f\in L^2(\mathbb{R}).$ Now to prove \begin{align} f*f=f. \end{align} Clearly, $f*f(x)=\int_{\mathbb{R}}f(t)f(x-t) dt$ exist pointwise due to Cauchy--Schwartz inequality. Now, Let's take , $h=F^{-1}(g^2)$, we will prove that $h=f*f$ a.e. \begin{align} h(x)&=\int_{\mathbb{R}} g^2(t) e^{ixt} dt a.e.\\ &=\int_{\mathbb{R}} F(f) F(f) e^{ixt} dt\\ &=\int_{\mathbb{R}} F(f)(t) F(\tau_{-x} f)(t) dt\\ &= \int_{\mathbb{R}} f(-t) f(t+x) dt\\ &=\int_{\mathbb{R}} f(t) f(x-t) dt\\ &=f*f(x), a.e. \end{align} So, $f*f=h=F^{-1}(g^2)=F^{-1}(g)=f$, since $g^2=g.$