Proof that the dirac delta function is a member of $H^{-1}(\mathbb{R})$

Question

I see this fact commonly stated but I haven't been able to find a formal proof of this fact anywhere.

Thanks.

Answer 1

Yes, this is an important fact to understand (including various proofs, a.k.a. "mechanisms"). It may often get "lost in the shuffle" of fancier things, or consigned to an exercise...

One proof proceeds from the characterization of with $H^{-1}(\mathbb R)$ as the (continuous) dual of $H^1(\mathbb R)$. Sobolev imbedding/inequality gives (a continuous inclusion) $H^{1/2+\varepsilon}(\mathbb R)\subset C^0(\mathbb)$, with the Frechet space structure on the latter. Then elementary estimates show that evaluation at $0$ is a continuous linear functional on $C^0(\mathbb R)$, so there we are. (Indeed, $\delta$ is in $H^{-1/2-\varepsilon}(\mathbb R)$ for every $\varepsilon>0$.)

Another proof uses the Fourier transform characterization of $L^2$-Sobolev spaces (and gives the more precise estimate of where $\delta$ lies). Namely, the Fourier transform of $\delta$ is computed (by various means, not only heuristic but also rigorous) to be $1$. Since $\int_{\mathbb R}|1|^2\cdot dx/(1+x^2)^s<+\infty$ for all real $s>{1/2+\varepsilon}$, we have $\delta$ in the corresponding Sobolev space.

Yes, there are some tricks that give simpler heuristic arguments than these, but these are somehow (to my thinking, anyway) the most honest explanations. Neither is trivial, indeed.