Let $f\in$$L^1(\mathbb{R})$ and $g(y)=\int_{\mathbb{R}}f(x)e^{-iyx}dx.$ Then if $f$ has compact support, $g=\hat f(y)$ can not have compact support unless $f \equiv0$. Similarly, let $\Omega_c^{1}(X)$ denote the space of continuous one-forms with compact support on a Riemann surface $X.$
Is there a generalization of the above theorem for differential forms, whereby if $\omega\in\Omega_c^{1}(X)$ has compact support, and $$\sigma(\xi)=\int_{\partial X}e^{-i\xi z}\omega$$ for $\omega=f(z)dz,$ then $\sigma(\xi)=\hat \omega(\xi)$ can not (necessarily) have compact support unless $\omega\equiv0?$
Proposition/Attempted Proof: Let $\pi$ be a diffeomorphism $\pi:\mathbb{R} \to X,$ such that $$\sigma(\xi)=\int_{\partial X}e^{-i\xi z}f(z)dz\equiv\int_{\pi(\mathbb{R})}e^{-i\xi \pi(x)}f(\pi(x))\pi^{\prime}(x)dx,$$ where $\pi(x)=z,\pi(y)=\xi.$ Similarly, let $t=Re(\pi)(x)=\frac{\pi+\bar \pi}{2}$ be a change of variables, then $$\sigma(\xi)=\int_{\mathbb{R}}e^{-i\xi t}f(t)\frac{dt}{dx}dx=\int_{\mathbb{R}}e^{-i\xi t}f(t)dt,$$ where $\pi(\mathbb{R})$ is transformed into $\mathbb{R}$ under $t=Re(\pi)(x).$ This reduces the above case into the less general $g(y)=\int_{\mathbb{R}}f(x)e^{-iyx}dx.$ However, there seems to be a logical fallacy in this proof.
Thanks in advance.
Proof: Let $\pi:\mathbb{R}\to\partial X$ be a homeomorphism, modulo some non-rectifiability condition on $\partial X.$ Similarly, let $\sigma(\chi)=\int_{\partial X}e^{-i\xi\chi}g(\xi)d\xi,$ for $\chi\in\mathbb{R}$ and $\omega=g(\xi)d\xi$ a differential 1-form on a Riemann surface $X,$ belonging to the space of continuous 1-forms $\Omega_{c}^1(X)$ with compact support on $X.$ Then $$\sigma(\chi)=\int_{\partial X}e^{-i\xi\chi}g(\xi)d\xi\equiv \int_{\pi(\mathbb{R})}e^{-i\pi(x)\chi}g(\pi(x))\pi^{\prime}(x)dx.$$ Hence let $t=T(\pi)=Re(\pi)(x)=\frac{\pi+\bar\pi}{2},$ whereby $T:X\to\mathbb{R},$ for $\mathbb{R}=Re(\partial X)\equiv Re(\pi(\mathbb{R})).$ We invoke the following theorem: Suppose that $X\subset V\subset\mathbb{R}^k,V$ is open, $T:V\to\mathbb{R}^k$ is continuous; $X$ is Lebesgue measurable, $T$ is one-to-one on $X,$ and $T$ is differentiable at every point of $X$;$m(T(V-X))=0.$ Then setting $Y=T(X),$ $$\int_Yfdm=\int_X(f\circ T)|J_T|dm$$ for every measurable $f:\mathbb{R}^k\to[0,\infty]$ and $|J_T(x)|=\Delta(T^{\prime}(x))=\frac{m(T(E))}{m(E)}.$ Therefore $$\int_{\pi(\mathbb{R})=\partial X}e^{-i\pi(x)\chi}g(\pi)\frac{d\pi}{dT}\frac{dT}{dx}dx\equiv\int_{T(X)=\mathbb{R}}e^{-ix\chi}g(x)\left(\frac{d\pi}{dT}\right)dx,$$ where $\frac{d\pi}{dT}=\frac{d\pi}{dRe\pi}=1,$ establishing an isometry. Consequently $$\sigma(\chi)=\int_{\mathbb{R}}e^{-ix\chi}g(x)dx=\int_{\mathbb{R}}e^{-it\chi}g(t)dt.$$ This can be extended to $\sigma(\phi)=\int_{\mathbb{R}}e^{-it\phi}g(t)dt$ for $\phi\in\mathbb{C}.$ If we assume the contrapositive, i.e., if $\sigma$ is zero on an open subset of $\mathbb{R},$ then it is also identically zero on $\mathbb{C}.$ Consequently, if $g(t)dt$ has compact support (zero on an open subset of $\mathbb{R}$), then $g(\xi)d\xi$ is also zero on an open subset of $X,$ by extension of the above integral $\int_{\partial X}e^{-i\xi\chi}g(\xi)d\xi=\int_{\mathbb{R}}e^{-it\chi}g(t)dt$ (Similarly, by this relation, the compact support of $g(\xi)d\xi$ on $X$ implies the compact support of $g(t)dt$ on $\mathbb{R}$), for $\pi:\mathbb{R}\to\partial X$; assuming $d\pi=d\xi=\pi_{*}$ with $\pi$ non-constant. Thus, if $\omega=g(\xi)d\xi$ has compact support, then $g(t)dt$ must have compact support. It follows that $\sigma(\chi)$ does not have compact support, i.e., the compact support of $\omega$ implies that $\sigma(\chi)$ can not have compact support, as required.