Why is dirac delta in $H^{-1/2-\epsilon}$ and why is multiplication not well-defined in that space?

Question

I read on this Sobolev space with negative index question that the dirac delta distribution belongs to $H^{-1/2-\epsilon}(\mathbb{R})$. I was trying to figure out why that is the case but it wasn't clear to me. I know that it acts on test functions by $\langle \delta, f\rangle = f(0)$ Although I don't see why the test functions necessarily need to be in $H^{1/2}$.

Finally, I also read that $\delta^2$ is not well defined as a distribution. How would I see this? Is it simply because if we apply $\delta$ to a function once, we get a value in $\mathbb{R}$. But then why not treat $f(0)$ like a constant function that you can then apply $\delta$ to again?

I can see the issue with naively defining $\langle \delta^2,f\rangle = f(0)^2$, because here linearity breaks down. But is there a general proof for why there's no good way to define $\delta^2$?

Answer 1

For the first question, this is due to the fact that $H^{s}(\mathbb{R}^n)$ is continuously embedded in $C^0(\mathbb{R}^n)$ if and only if $s>n/2$ which for $n=1$ gives $s>1/2$.

For the second one, you cannot define consistently a multiplication between distribution, there were a lot of attempts for that in the eighties (see works of JF Colombeau for instance) but no general framework found as far as I know.

Moreover it seems that in your question you are messing between multiplication and composition. You cannot compose $\delta$ by itself obviously because its image space $\mathbb{R}$ is not its domain space (continuous functions).

The natural multiplications of distribution is rather convolution, and it turns out that $\delta$ is the neutral element for this operation, so that $\delta\star\delta=\delta$.

Answer 2

As a complement to Emmanuel nice answer, you can also compute directly the $H^s$ norm of the Dirac delta since $\mathcal F(\delta_0) = 1$, you get $$ \|\delta_0\|_{H^{-s}}^2 = \int_{\Bbb R^d} |\mathcal F(\delta_0)|^2\left(1+|x|^2\right)^{-s} \\= \int_{\Bbb R^d} \frac{\mathrm d x}{\left(1+|x|^2\right)^s} $$ and this integral converges if and only if $s > d/2$ (so $s>1/2$ if $d=1$).