1 in Sobolev spaces H^s

Question

Following Rauch's Partial Differential Equations, he defines the generalization of the Sobolev Spaces for any $s\in\mathbb{R} $ as $$H^s(\mathbb{R}^n)=\{u\in S' | (1 + |x|^2)^{s/2}\widehat{u}\in L^2(\mathbb{R}^n) \} $$ Then, he gives the usual examples: Dirac's Delta $\delta$ belongs to $H^s$ for $s<-n/2$, and that's easy to see because for any test function $\phi$ $$\langle \widehat{\delta}, \phi \rangle = \langle \delta, \widehat{\phi} \rangle = \widehat{\phi}(0) = \int_{\mathbb{R}^n} (2\pi)^{-n/2} \phi(x) dx = \langle(2\pi)^{-n/2},\phi \rangle$$

So $\widehat{\delta} = (2\pi)^{-n/2} $. Ergo, it's enough to check if the integral $$\int_{\mathbb{R}^n}(1 + |x|^2)^{s} dx$$ converges. This is gotten by passing to spheric coordinates, leaving the convergence depending on the radius (only 1D), because all the other dimensions become nice bounded integrable functions whose domain doesn't depend of the radius. So, making the quotient comparation with $r^{2s}$ for $r$ big enough leaves de desired result.

Well, that's when the feel of understanding stuff goes to hell. Example 2 is 1 back, saying that $1$ isn't in any $H^s$. Here is where the existential dilemma comes. As we know, for a test funtion $\phi$ $$\langle \widehat{1}, \phi \rangle = \langle 1, \widehat{\phi} \rangle = \int_{\mathbb{R}^n} \widehat{\phi}(x) dx = (2\pi)^{n/2} \phi(0) = \langle (2\pi)^{n/2}\delta , \phi \rangle$$ So $ \widehat{1} = (2\pi)^{n/2}\delta$, but how the heck I'm supposed to check out if $(2\pi)^{n/2}\delta(x)(1+|x|^2)^{s/2}$ is in $L^2$? First than anything, does $\delta^2$ has any sense?

Thank you beforehand, and sorry by my messy English, I'm Chilean

Answer 1

First than anything, does $\delta^2$ has any sense?

It does not.

how the heck I'm supposed to check out if $(2\pi)^{n/2}\delta (x)(1+|x|^2)^{s/2}$ is in $L^2$?

By duality. This was executive summary; details are below.

You have a distribution $T$, and want to know if there is an $L^2$ function $g$ such that $$T(\phi)=\int_{\mathbb R^n} \phi g \tag{1}$$ for all test functions $\phi$. If (1) holds, then $|T(\phi)|\le \|\phi\|_{L^2}\|g\|_{L^2} $ by Cauchy-Schwarz. Put simply, $T$ is bounded in the $L^2$ norm: $$|T(\phi)|\le C\|\phi\|_{L^2} \tag{2}$$ with $C$ independent of $\phi$. (And conversely, if (2) holds, then $T$ extends to a bounded functional on $L^2$, and Riesz representation delivers $g$ as in (1).)

For your particular $T$ (which, by the way, is simply $ (2\pi)^{n/2}\delta$) you have $T(\phi)=(2\pi)^{n/2}$ whenever $\phi(0)=1$. Since a test function with $\phi(0)=1$ can have arbitrarily small $L^2$ norm, (2) fails.

Answer 2

Ok, here I am answering myself xD Thanks to icurays1, the right neurons made synapse finally xD Let's suppose that $(2\pi)^{n/2}\delta(x)(1+|x|^2)^{s/2}$ is in $L^2$. If that's true, the inverse fourier Transform should also be in $L^2$. But: $$\mathcal{F}^{-1}[(2\pi)^{n/2}\delta(x)(1+|x|^2)^{s/2}](\lambda) = \int_{\mathbb{R}^n} e^{i \lambda \cdot x}(2\pi)^{n/2}\delta(x)(1+|x|^2)^{s/2} dx = (2\pi)^{n/2}$$ Which obviously ain't on $L^2$, for any $s\in\mathbb{R}$. Dang it, i'm feeling dumb, sorry for the trouble, and thanks for your time, guys.