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?
