I know that delta distribution $\delta : \mathcal S (\mathbf R) \to \mathbf C$ is continuous with usual seminorm and here.
I am interested in its continuity with dual-space $H^{-1}(\Omega)$ of $H_{0}^{1}(\Omega)$, which I think is based on $L_2(I)$ norm in its denominator:
\begin{equation} \lVert f \rVert_{L_{2}(\Omega)} = \sup_{ \substack{v \in L_{2}(\Omega) \\ v \neq 0 } } \frac{ |(f,v)|}{ \lVert v \rVert_{L_{2}(\Omega)}} \end{equation}
We define the space of square integrable functions on $I$: \begin{equation} L_{2}(I) = \{ v: v \text{ is defined on } I \text{ and } \int_{I} v^{2} dx < \infty \}. \end{equation}
I want also that the delta distribution is differentiable, which do by duality.
Is delta distribution continuous and differentiable with dual space norm?
Let $\varphi$ be a test function (i.e. $C^\infty$ with compact support) with $\varphi(0)=1$ and set $\varphi_n(x)=\varphi(nx)$. Then $\varphi_n$ is a test function and you have using the substitution $nx\mapsto t$ $$ ||\varphi_n||_{L^2} = \Bigl( \int |\varphi(nx)|^2 dx \Bigr)^\frac{1}{2} = \Bigl(\frac{1}{n} \int |\varphi(t)|^2 dt \Bigr)^\frac{1}{2}\longrightarrow 0 \,\text{ as }\ n\to\infty. $$ On the other hand you have $\delta(\varphi_n) = \varphi_n(0) = \varphi(0) = 1$ for every $n$. This proves the non-continuity of $\delta$ with respect to $L^2$-norm.