Consider the following transform $T:g\mapsto f$, where $$ f(x)=\int_{0}^{\infty} \exp\left\{-(k_t * g)(x)\right\}\,dt $$ where $k_t$ is the "tent" function $$ k_t(u)=\begin{cases} t-|u|,&|u|\leq t,\\ 0,&|u|>t. \end{cases} $$ Assume $f\geq 0$ and $g>0$ are $C^\infty(\mathbb{R})$. How does one tackle the question of whether this transform is injective?
Some thoughts: We want to show that $T(g_1)=T(g_2)$ implies $g_1=g_2$, or find an example where it does not. Could we perhaps use the Fourier transform and the convolution theorem? Any ideas would be appreciated.
My attempt: The tent function $k_t(u)$ has Fourier transform $$ \mathcal{F}(k_t)(\omega) = \frac{4 \sin^2(\omega t/2)}{\omega^2}. $$ Then, by the convolution theorem, the Fourier transform of $k_t * g$ is $$ \mathcal{F}(k_t * g)(\omega) = \mathcal{F}(k_t)(\omega) \cdot \mathcal{F}(g)(\omega). $$ The zeros of $\mathcal{F}(k_t)$ occur at $\omega = \frac{2\pi k}{t}$ for $k \in \mathbb{Z} \setminus \{0\}$. As $t$ varies over $\mathbb{R}^+$, these zeros cover all non-zero frequencies. If $k_t * (g_1 - g_2) = 0$ for all $t$, then $\mathcal{F}(g_1 - g_2)$ must vanish everywhere, implying $g_1 = g_2$.
Suppose $\int_{0}^{\infty} \exp\{ -(k_t * g_1)(x) \} \, dt = \int_{0}^{\infty} \exp\{-(k_t * g_2)(x)\} \, dt$ for all $x$. By the mean value theorem for integrals, this implies $$ \exp\{ -(k_t * g_1)(x) \} = \exp\{ -(k_t * g_2)(x) \} \quad \forall t, x. $$ Taking logarithms gives $k_t * (g_1 - g_2) = 0$ for all $t, x$, which forces $g_1 = g_2$. Hence, $T$ is injective. Is this correct?
