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The delta function is well known to be in $H^s(\mathbb{R})$ for $s<-\frac{1}{2}$, but not in $H^{-\frac{1}{2}}(\mathbb{R})$. Given that $H^{-\frac{1}{2}}$ is the dual of $H^{\frac{1}{2}}$, there must be some $f\in H^{\frac{1}{2}}(\mathbb{R})$ such that $<\delta,f>$ is undefined, since otherwise $\delta$ would be in $H^{-\frac{1}{2}}(\mathbb{R})$ by definition.

I would like an example of such an $f$. I am under the possibly mistaken impression that elements of $H^{\frac{1}{2}}(\mathbb{R})$ have well-defined values at points by the trace theorem.

Zorgoth
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  • Have you checked continuity? – Anthony Carapetis May 09 '17 at 22:41
  • Oh. That would be it. – Zorgoth May 09 '17 at 22:55
  • Should that first inequality be $s<-1/2$? – Jonas Dahlbæk May 09 '17 at 23:10
  • As far as I know, a function from $H^{1/2}(\mathbb R)$ can have a singularity. Can you give a pointer to your 'trace theorem'? – gerw May 10 '17 at 13:18
  • Ok, now I see two problems with my problem.

    The trace theorem states that for a suitably nice domain $\Omega\subset\mathbb{R}^n$ and certain values of $s$, $u\in H^s(\Omega)\implies u|_{\partial\Omega}\in H^{s-\frac{1}{2}}(\partial\Omega)$

    Here's an example:

    http://www.ams.org/journals/proc/1996-124-02/S0002-9939-96-03132-2/S0002-9939-96-03132-2.pdf

    I was thinking that the Trace theorem held when $s=\frac{1}{2}$. I think that was a mistake ($s$ must be bigger than $\frac{1}{2}$). And even if delta can be shown to be defined (which I think it can't), it would need to be bounded.

     – Zorgoth May 10 '17 at 22:38
  • Yes, Jonas. I fixed it. Thanks. – Zorgoth May 10 '17 at 22:39

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Being in $H^{1/2}(\mathbb{R})$ allows for singularities that are milder than logarithmic: such as $\log\log|x|$ near $0$ (cut off to create a function with compact support). Perhaps the best way to see this is to use the fact that $H^{1/2}(\mathbb{R})$ is the trace space of $H^1(\mathbb{R}_+^2)$, and to verify that $u(x,y)=\log\log \sqrt{x^2+y^2}$ is in $H^1$ near the origin. This is done here.