What exactly is the $H^{-1/2}$ space?
Definition for $H^{1/2}$ given e.g. here.
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$Y = \widehat{H^{1/2}(\mathbb{R})}$ is the Hilbert space of functions $f \in L^2(\mathbb{R})$ with the norm $$\|f\|_Y^2 = \int_{-\infty}^\infty (1+|x|) |f(x)|^2dx$$ ie. the inner product $$\langle f, g \rangle_Y = \int_{-\infty}^\infty (1+|x|) f(x)\overline{g(x)}dx$$
Its strong dual is $Y^*=\widehat{H^{1/2}(\mathbb{R})}^*$, the Hilbert space of functions with the norm $$\|g\|_{Y^*}^2 =\sup_{f \in Y} \frac{|\langle f, g \rangle_Y|^2}{\|f\|_Y^2}= \int_{-\infty}^\infty \frac{1}{1+|x|} |g(x)|^2dx$$ and inner product $$\langle g_1,g_2 \rangle_{Y^*} = \int_{-\infty}^\infty \frac{1}{1+|x|} g_1(x)\overline{g_2(x)}dx $$
Then $X=H^{1/2}(\mathbb{R})$ is the inverse Fourier transform of $Y$ and $X^*=H^{-1/2}(\mathbb{R})=H^{1/2}(\mathbb{R})^*$ is its strong dual, the inverse Fourier transform of $Y^*$.
We can also define those things in term of fractional derivatives (some self-adjoint and closed unbounded operators).
Restricting the norm to $\Omega$ we obtain $H^{\pm 1/2} (\Omega)$.
reuns
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I've not met the term strong dual before. Just dual. – mavavilj Nov 23 '17 at 03:58
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Also, what does $\overline{\cdot}$ denote? – mavavilj Nov 23 '17 at 03:59
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Strong dual (of a Banach space) just means space of continuous linear functionals with the operator norm. For Hilbert spaces we always consider the strong dual (an Hilbert space too) – reuns Nov 23 '17 at 04:01
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Also, what do you denote with \widehat over the $H$? – mavavilj Nov 23 '17 at 22:15
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fourier transform.. – reuns Nov 24 '17 at 05:29