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Consider the following convolution

$$ f(x)=(k * g)(x)=\int_{-\infty}^{\infty}k(t)g(x-t) \,{\rm d} t $$

where $g>0$, and $k$ is the "tent" function

$$ k(t) = (1 - |t|)\chi_{[-1,1]}(t) $$

where $\chi$ is the characteristic function. Is this transform invertible? That is, can I write $g$ in terms of $f$? If not, is it at least injective?

My attempt: Define $T(g) := k * g$. By the convolution theorem, \begin{equation} \widehat{T(g)}(\omega) = \widehat{k * g}(\omega) = \hat{k}(\omega)\hat{g}(\omega). \end{equation} where the hat represents the Fourier transform. A direct calculation shows \begin{equation} \hat{k}(\omega) = \left(\frac{\sin(\omega/2)}{\omega/2}\right)^2, \end{equation} which vanishes whenever $\omega = 2\pi n$ for any nonzero integer $n$.

Non-invertibility: If we attempt to recover $g$ from $f = k * g$ by \begin{equation} \hat{g}(\omega) = \frac{\hat{f}(\omega)}{\hat{k}(\omega)}, \end{equation} we require $\hat{k}(\omega) \neq 0$ for all $\omega$. However, since $\hat{k}(\omega) = 0$ at infinitely many points, we cannot divide by $\hat{k}(\omega)$ globally. Hence $T$ is not invertible.

Non-injectivity: If $T(g) = 0$, then \begin{equation} \hat{k}(\omega)\hat{g}(\omega) = 0 \quad \text{for all } \omega. \end{equation} Since $\hat{k}(\omega) = 0$ on $\{\omega = 2\pi n \mid n \neq 0\}$, any $\hat{g}$ supported in those zeros will satisfy $\hat{k}\hat{g} = 0$ yet $g \not\equiv 0$. Thus $T(g) = 0$ does not imply $g = 0$, so $T$ is not injective. I am wondering whether this is true, or whether it holds in the $g>0$ case. Any ideas?

Rodrigo de Azevedo
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sam wolfe
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    Just an advice: problems of this kind are known in the literature under the locution "deconvolution problems". You may find something useful by just googling the term "deconvolution". – Daniele Tampieri Feb 05 '25 at 10:55
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    Thank you! If you have any reference on how to tackle a specific problem like mine it would be great. – sam wolfe Feb 05 '25 at 11:37

2 Answers2

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First of all, to have this question make sense you need to specify what space of functions you're dealing with. I'll assume you're working in $L^2(\mathbb R)$.

The fact that $\hat{k}(\omega) = 0$ at some points does not make $T$ non-injective, because members of $L^2$ are only equivalence classes of functions under equality almost everywhere: single points, or even sets of measure $0$, don't matter. A member of $L^2$ that is $0$ except at points $\omega = 2 \pi n$ is $0$, and its inverse Fourier transform is $0$. So $T$ (as an operator on $L^2$) is injective. On the other hand, it is not surjective: e.g. in order to have $\hat{k} \hat{g}$ be bounded away from $0$ in a neighbourhood of $2 \pi n$, $\hat{g}$ must have a singularity at $2 \pi n$ that is not square-integrable.

Robert Israel
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The short answer is no. Consider a $1$-period function $g$ on $\mathbb R$, you have $\chi_{[-1/2,1/2]}\ast g(x)=\int_0^1g$ is a constant function. Meanwhile $k=\chi_{[-1/2,1/2]}\ast\chi_{[-1/2,1/2]}$.

However consider your another question this is not the end of the world, as you can strengthen the assumption to be $k_t\ast g=0$ for all $t>0$.

Liding Yao
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