This is a question about the action of the Dirac delta on Sobolev spaces $H^s(\mathbb{R}^d) = W^{s,2}(\mathbb{R}^d)$.
We know that $\delta(\underline{x})\in H^s(\mathbb{R}^d)$ for $s<-d/2$.
In 2D, consider $v \in H^2(\mathbb{R}^2)$. Since $\delta(\underline{x})\in H^{-2}(\mathbb{R}^2)$ (the dual of $H^2(\mathbb{R}^2)$), the Dirac delta acts on $v$ as $$\int_{\mathbb{R}^2} \delta(\underline{x}) v(\underline{x}) \; \mathrm{d}\underline{x} = v(\underline{0}).$$ Furthermore, in 2D, we have the Sobolev embedding $H^2(\mathbb{R}^2) \subset C^0(\mathbb{R}^2)$, so the value of $v$ at $\underline{0}$ is well-defined.
However in 3D I am a bit confused: Take $v \in H^2(\mathbb{R}^3)$, and we still have $\delta(\underline{x})\in H^{-2}(\mathbb{R}^3)$. The action of $\delta$ on $v$ should be well-defined as above. However, in 3D we know that $H^2(\mathbb{R}^3) \not\subset C^0(\mathbb{R}^3)$, so I can't see how the answer of $v(\underline{0})$ is well-defined as we shouldn't be able to talk about point values of $v$.
Thanks.
I think the mistake is that $$ H^2(\mathbb R^3) \subset C^0(\mathbb R^3) $$ is true.
Using the Sobolev embedding theorem, we can see that $H^2(\mathbb R^3) =W^{2,2}(\mathbb R^3)\subset W^{1,6}(\mathbb R^3)$.
Then using the part of the theorem that uses Holder spaces, we have $ W^{1,6}(\mathbb R^3) \subset C^{0,\tfrac12}(\mathbb R^3). $