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Added an example of an invertible element that does not have support $\{0\}$.
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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 $\delta_n \ast \delta_{-n}=\delta_0$.)

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

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

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

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