$GL_1(\mathcal{E}'(\mathbb{R}))$ open in $\mathcal{E}'(\mathbb{R})$?

Question

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 $\delta_n \ast \delta_{-n}=\delta_0$.)

Answer 1

Perhaps Jochen Wengenroth's comments already give the answer, but here's a direct argument.

By Paley-Wiener 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 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.