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Let $p,\mu,\nu$ be probability density functions on $\mathbb{R}$ such that $$ \int_{\mathbb{R}}p(y-x) \nu(y) \, dy=\mu(x). $$ Now, consider the operator $T:L^2(\mu)\to L^2(\nu)$ such that $$ Tf=f*p.$$ Note that $f*p(x)$ is well defined for almost every $x$ as using the Jensen inequality one gets $$ \int_{\mathbb{R}} (|f|*p)^2(x)\nu(x) \, dx \le \int_{\mathbb{R}}f^2(y)\int_{\mathbb{R}}p(x-y)\nu(x) \, dx \, dy =\int_{\mathbb{R}} f^2(y)\mu(y)<\infty. $$

Edit

Let us assume that $p,\mu,\nu$ has non-vanishing (everywhere) Fourier transforms. Let us also assume that $p,\mu,\nu$ are infinitely divisible densities.

Question: Can we prove that $T$ is injective?

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    $\begingroup$ “Fourier transform of $p$ is non-zero almost everywhere“? $\endgroup$
    – Fedor Petrov
    Commented May 15 at 5:21
  • $\begingroup$ @FedorPetrov: what if $\nu$ has a non-trivial zero-set? It is possible that $f*p$ is not zero as a function but is equivalent to $0$ as an element of $L^2(\nu)$. $\endgroup$
    – Willie Wong
    Commented May 15 at 7:51
  • $\begingroup$ Well, in "almost everywhere" a measure must be specified, but something like that should be true $\endgroup$
    – Fedor Petrov
    Commented May 15 at 9:01
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    $\begingroup$ Condition $p*f \neq 0$ for any $0\neq f \in L_2(\mu)$ is equivalent to $p*g \neq 0$ for any $0 \neq g \in V=\frac{\chi_{\mu \neq 0}}{\sqrt{\mu}} L_2(\lambda)$, where $\lambda$ is a Lebesgue measure. To pass to the Fourier transform $\hat{p} \cdot \hat{g}$ we should be able to multiply Fourier images. So one should make sense of $\hat{p} \cdot \hat{g} $ for any $0\neq g \in F(V)$ to make use of Fourier transform. $\endgroup$
    – plllnt
    Commented May 15 at 11:37
  • $\begingroup$ Additionally, if $\mu = (1+|x|^2)^{-1}$, then $g$ might be equal to $e^{ikx}$, which would make $\hat{g}$ proportional to a delta function at any point $k$. If $\hat{p}$ is smooth, one can see that almost everywhere invertibility is not enough. $\endgroup$
    – plllnt
    Commented May 15 at 11:38

2 Answers 2

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Let $V(\mu)=\sqrt{\mu} L_2\subset L_1$ (with Lebesgue measure) and $V'(\mu)$ be a dual of $V'(\mu)$ with respect to pairing $(f,v) = \int f v dx$. The set $V'(\mu)$ can be thought of as $\frac{\chi_{\operatorname{ess-sup}\mu}}{\sqrt{\mu}} L_2$ which we can associate with $L_2(\mu)$. We can define Fourier transform on $V'(\mu)$ the usual way: $$ (F(f), v) = (f, F(v)),\quad v\in F^{-1}(V(\mu)). $$ We have $F(V'(\mu)) = (F(V(\mu))'$.

Note that $p\in L_1$, so $F(p)\in C_0(\mathbb{R})$ is a regular function. But to have an equality $F(p*g)=F(p)F(g)$ we need $F(p)w \in F^{-1}(V(\nu))$ for any $w \in F^{-1}(V(\nu))$, or equivalently, $\tilde{p}*v \in V(\mu)$ for any $v \in V(\nu)$, here $\tilde{p}(x)=p(-x)$, since then we would have $$ \begin{align} &(F(p*g), w) = (p*g, F(w)) = (g, \tilde{p}*F(w)) = (g,FF^{-1}(\tilde{p}*F(w))) \\ & = (F(g), F(p)w) = (F(g)F(p),w). \end{align} $$ As you noted, due to the fact that $$ \int p(y-x)\nu(y)dy = \int p(-y)\nu(x-y)dy\mu(x) $$ we get $p*g \in V'(\nu)$ for any $g \in V'(\mu)$. Using the same equality we can obtain $$ \begin{align} |(v*\tilde{p})(x)|^2 &\leq \left( \int \sqrt{\nu(x-y)} \sqrt{p(-y)}|u(x-y)| \sqrt{p(-y)} dy \right)^2 \\ &\leq \left(\int p(y-x) \nu(y)dy\right) \left(\int|u(x-y)|^2 p(-y) dy\right) \\ &= \mu(x) |s(x)|^2, \end{align} \label{1}\tag{$*$} $$ where $v=\sqrt{\nu} u \in V(\nu)$, $s \in L_2$. Therefore $v*p \in V(\mu)$.

Now, in order to divide by $F(p)$ in $F(p)F(g)=0$, in addition to $F(p)(\xi)\neq 0$ at any point $\xi$ we need $\frac{1}{F(p)}w \in F^{-1}(V(\nu))$ for any $w \in F^{-1}(V(\mu ))$, or equivalently $F(\frac{1}{F(p)})*v \in V(\nu)$ for any $v \in V(\mu)$. From $\int p(y-x)\nu(y)dy =\mu(x)$ we have $\frac{1}{F(\tilde{p})}F(\mu) = F(\nu)$. Note that $F^{-1}(\frac{1}{F(\tilde{p})}) = F(\frac{1}{F^{-1}(\tilde{p})}) = F(\frac{1}{F(p)})$. Therefore $$ \int F(\frac{1}{F(p)})(x-y) \mu(y) dy = \nu(x). $$ Similar to inequality \eqref{1} for $\tilde{p}$ replaced with $F(\frac{1}{F(p)})$ and $\nu$ replaced with $\mu$ we get $F(\frac{1}{F(p)}) *v \in V(\nu)$ for any $v \in V(\mu)$. So indeed $\frac{1}{F(p)}w \in F^{-1}(V(\nu))$ for any $w \in F^{-1}(V(\mu ))$. Thus, $T$ is injective if $F(p)(\xi)\neq 0$ at any point.

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  • $\begingroup$ @ChristianRemling You are right, thanks for pointing out. It should have been $\sqrt{\mu} L_2 \subset L_1$. This condition is needed to define Fourier transform on 'test functions' $V(\mu)$, and it is well defined on $L_1$. I'll edit accordingly. $\endgroup$
    – plllnt
    Commented May 18 at 23:37
  • $\begingroup$ Thank you. I have one more concern, which is perhaps really the same as my comment above on the OP's question: If I understand your argument correctly, you assume that $F(p)F(g)=0$ and then conclude that $F(g)=0$. But is it actually clear that $F(p*g)=F(p)F(g)$? More to the point, the convolution cannot really be made sense of directly for arbitrary $g\in V'(\mu)$ if $\mu$ vanishes to high order somewhere because then $V'(\mu)$ contains functions $g$ that are very far from being locally integrable. $\endgroup$
    – Christian Remling
    Commented May 18 at 23:57
  • $\begingroup$ @ChristianRemling First of all, OP already showed that $p*g$ is well defined for any $g \in V'(\mu)$. Even though $g$ can be highly singular, it is still a function, and multiplying it by $p(y-x)$ makes sense and the whole expression is integrable which is shown in the inequality in OP's question. $\endgroup$
    – plllnt
    Commented May 19 at 0:28
  • $\begingroup$ @ChristianRemling Secondly, to prove that $F(p*g)=F(p)F(g)$ is what this whole thing is about. In order for this expression to be well defined we need every step in $(F(p*g), F^{-1}v) = (p*g, v) = (g, \tilde{p}*v)$ $= (g, FF^{-1} (\tilde{p}*v) =$ $(F(g), F(p) F^{-1}(v)) = (F(g)F(p), F^{-1}(v))$ to make sense, here $\tilde{p}(x)= p(-x)$ (I may be missing a sign). I'll edit an answer to make that point more clear, and correct some misprints and signs in the arguments too. $\endgroup$
    – plllnt
    Commented May 19 at 0:35
  • $\begingroup$ @ChristianRemling And sorry for an off-topic, why do I not see your previous comments? Is this a bug? $\endgroup$
    – plllnt
    Commented May 19 at 0:37
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Let $p$, $\mu$, and $\nu$ be infinitely divisible probability density functions on $\mathbb{R}$ satisfying the given convolution equation. Consider the operator $T: L^2(\mu) \to L^2(\nu)$ defined by $Tf = f * p$. To prove injectivity, assume $Tf = 0$. Taking the Fourier transform, we have $\mathcal{F}(Tf) = \mathcal{F}(f) \cdot \mathcal{F}(p) = 0$. Since $p$ is a probability density, $\mathcal{F}(p)(0) = 1$, and we assume that $\mathcal{F}(p)$ is non-vanishing, so $\mathcal{F}(f) = 0$ almost everywhere.

Now, since $p$, $\mu$, and $\nu$ are infinitely divisible, there exist continuous functions $\psi$, $\phi$, and $\chi$ such that $\mathcal{F}(p) = e^{\psi}$, $\mathcal{F}(\nu) = e^{\phi}$, and $\mathcal{F}(\mu) = e^{\chi}$. The convolution equation's Fourier transform yields $e^{-\psi(-\xi)} \cdot e^{\phi(\xi)} = e^{\chi(\xi)}$. Taking logarithms, we obtain $-\psi(-\xi) + \phi(\xi) = \chi(\xi)$.

Since $\mathcal{F}(Tf) = \mathcal{F}(f) \cdot e^{\psi} = 0$, if there exists a $\xi$ such that $\mathcal{F}(f)(\xi) \neq 0$, then $e^{\psi(\xi)} = 0$, which implies that $\psi(\xi)$ is undefined, contradicting the continuity of $\psi$. Therefore, $\mathcal{F}(f)(\xi)$ must be zero for all $\xi$. By the injectivity of the Fourier transform, we conclude that $f = 0$ everywhere, proving the injectivity of the operator $T$.

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  • $\begingroup$ the Fourier transform of $f$ may not be defined. Note that $f$ belongs to a weighted $L^2$ space, so $f$ may not belong to $L^1(\mathbb{R})$ or $L^2(\mathbb{R})$. $\endgroup$
    – Ribhu
    Commented May 17 at 15:18

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