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I prooved that the Dirac distribution $\delta_{0}$ is in the Sobolev space $H^{s}\left(\mathbf{R}^{n}\right)=\left\{f \in \mathcal{S}^{\prime}\left(\mathbf{R}^{n}\right)\left|\left(1+|\xi|^{2}\right)^{s / 2} \mathcal{F} f \in L^{2}\left(\mathbf{R}^{n}\right)\right\}\right.$ for every $s<-n / 2$

but I steel wrestling to proof that the Heaviside distribution $H$

$\forall x \in \mathbb{R}, H(x)=\left\{\begin{array}{lll}{0} & {\text { si }} & {x<0} \\ {1} & {\text { si }} & {x \geq 0}\end{array}\right.$

Doesn't belong to any Sobolev space $H^{s}(\mathbb{R})$, could you elaborate on that?

Thanks in advance!

Almendrof66
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    I take it your definition of $H^s$ is the space of distributions $f$ such that $(1+|\xi|)^s \hat{f}(\xi)\in L^2,$ where $\hat{\cdot}$ denotes the Fourier transform? – WoolierThanThou Jan 08 '20 at 09:39
  • $H^{s}\left(\mathbf{R}^{N}\right)=\left{f \in \mathcal{S}^{\prime}\left(\mathbf{R}^{n}\right)\left|\left(1+|\xi|^{2}\right)^{s / 2} \hat{f} \in L^{2}\left(\mathbf{R}^{N}\right)\right}\right.$ where $\mathcal{S}^{\prime}\left(\mathbf{R}^{n}\right)$ the space of tempered distributions – Almendrof66 Jan 08 '20 at 09:46
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    See this answer: https://math.stackexchange.com/a/1186815/686397. The Fourier Transform of $H$ is not a function, and thus, multiplying it by $(1+|\xi|^2)^{s/2}$ will never yield an $L^2$ function. – WoolierThanThou Jan 08 '20 at 09:51
  • As I understood from the answers on the Fourier transform of the Heaviside distribution is that it exists, and it's expressed by $H(\xi) = \frac{1}{i\xi} + \pi \delta(\xi)$ see for example https://math.stackexchange.com/questions/1337481/representation-of-heaviside-functions-fourier-transform, so the proof based on the fact that the Fourier Transform of H doesn't exist couldn't still valid – Almendrof66 Jan 08 '20 at 13:07
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    I din't say it didn't exist. I'm saying it's not a function. $\pi (1+|\xi|^2)^{s/2} \delta$ is never in $L^2$ because it's a measure singular to the Lesbegue measure. – WoolierThanThou Jan 08 '20 at 13:09
  • Fair enough, I apologize, I was confused. Now a get your answer – Almendrof66 Jan 08 '20 at 13:13
  • Let me share with you a proof of a friend mine: In one hand $H' = \delta \rightarrow \xi\ \hat{H}=-i \rightarrow$ $ \hat{H}\in L^2(\mathbb{R*})$. On the other hand, let's suppose that $H \in H^{s}(\mathbb{R})$ so $\hat{H}\in L^2_{loc} \rightarrow \hat{H}\in L^2(|\xi| \leq 1) \rightarrow$ $\hat{H} \in L^{2}(\mathbb{R}) \rightarrow$ $H \in L^{2}(\mathbb{R})$ absurd. I'll be very glad to read your comment on that [Is that correct if we start from considering that the Fourier Transform of H is correct] – Almendrof66 Jan 08 '20 at 13:50
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    I mean... the above proof relies on knowledge about $\hat{H}$, so I don't see much of a point? I mean... $\hat{H}$ clearly is not even a function, so why go through these extra steps? The proof is correct, though, as far as I can see. – WoolierThanThou Jan 08 '20 at 13:59

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