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.
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.