Let $\mathcal{E}'(\mathbb{R})$ be algebra of all compactly supported distributions on $\mathbb{R}$, equipped with the strong dual topology $\beta(\mathcal{E}',\mathcal{E})$, and with the usual operations of addition and convolution.


Is the set ${{\textrm{GL}}}\_1(\mathcal{E}'(\mathbb{R}))$ of invertible elements open in $\mathcal{E}'(\mathbb{R})$?


(As an example of an element in ${{\textrm{GL}}}\_1(\mathcal{E}'(\mathbb{R}))$ which does not have support $\{0\}$, we have that $\delta\_n$ belongs to ${{\textrm{GL}}}\_1(\mathcal{E}'(\mathbb{R}))$ 
because $\delta\_n \ast \delta\_{-n}=\delta\_0$.)



$GL_1(\mathcal{E}'(\mathbb{R}))$ open in $\mathcal{E}'(\mathbb{R})$?
Perhaps Jochen Wengenroth's comments already give the answer, but here's a direct argument.


By [Paley-Wiener](https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem) for distributions, the Fourier transform $\tilde{\Delta}(k)$ of a distribution of compact support $\Delta(x) \in \mathcal{E}'(\mathbb{R})$ is entire, of exponential type, with at most asymptotic polynomial growth on the real axis. Since convolutions become products under the Fourier transform, $\Delta(x) \in \mathrm{GL}\_1(\mathcal{E}'(\mathbb{R}))$ only if $\tilde{\Delta}(k)$ is pointwise invertible and $\tilde{\Delta}(k)^{-1}$ has the same analytic properties. But by the [Hadamard factorization theorem](https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem) this can only be if $\tilde{\Delta}(k) = e^{i k c\_1 + c\_0}$, with $c\_1 \in \mathbb{R}$, $c\_0\in \mathbb{C}$. So indeed, $\mathrm{GL}\_1(\mathcal{E}'(\mathbb{R}))$ consists only of scaled translates of $\delta\_0$.


I'm no big expert on the topology of $\mathcal{E}'(\mathbb{R})$, but I suspect this set is too small to be open.


