For $s>0 $ define $H^s(\mathbb{R}^n)=\lbrace u\in L^2(\mathbb{R}^n): (1+\vert y\vert^2)^\frac{s}{2}\hat{u}\in L^2(\mathbb{R}^n)\rbrace$. This is a Hilbert space with the inner product given by
$$\langle u,v\rangle=\int_{\mathbb{R}^n}(1+\vert y\vert^2)^s \hat{u}(y)\overline{\hat{v}(y)} dy.$$
This definition consist of complex functions. I was wondering if the restriction to real functions is a Hilbert space as well, but I can't see why $\langle u,v\rangle$ should be real valued, if $u$ and $v$ are.
A follow-up question that arises: How about the case for $s\in\mathbb{R}$, where one uses tempered distibutions in order to define $H^s(\mathbb{R}^n)$. In this case, I don't even really know what a real tempered distribution should be: As the name suggest it should be real valued. As a real $L^p$ functions $u\in L^p(\mathbb{R}^n)$ can be interpreted as tempered distributions, the resulting tempered distribution should be real valued. But I can choose a complex function $\varphi\in \mathcal{S}$, such that $\int_{\mathbb{R}^n}\varphi u \,dx$ will be complex as well. So should a real tempered distribution be only real valued on real test functions?! Is there some common approach? What happens to the restricted inner product? The Fourier transform of tempered distribution is defined by transposition. But the FT of the Schwartz functions will be complex, so it doesn't feel right to restict the distributions. Maybe the real numbers just aren't the right setting?!
This whole question came up only because I was looking for real solutions to a PDE and thought I could maybe restrict the space $H^s$.
Best Answer
The restriction certainly makes sense. If $u$ is real-valued then $\hat{u}(y)=\overline{\hat{u}(-y)}$ for all $y \in \mathbb{R}$ (I am sticking to one dimension for convenience). We then have \begin{align} &\int_{\mathbb{R}}(1+ y^2)^s\hat{u}(y)\overline{\hat{v}(y)} \, \mathrm{d}y \\ =& \int_{0}^\infty (1+ y^2)^s\left(\hat{u}(y)\overline{\hat{v}(y)} + \hat u(-y) \overline{\hat v(-y)}\right) \, \mathrm{d}y \\ =& \int_{0}^\infty (1+ y^2)^s\left(\hat{u}(y)\hat{v}(-y) + \overline{\hat u(y)} \,\overline{\hat v(-y)}\right) \, \mathrm{d}y \\ =& \int_{0}^\infty (1+ y^2)^s\left(\mathfrak{Re}(\hat{u}(y)\hat{v}(-y)) \right) \, \mathrm{d}y \, , \end{align} which is clearly real.
Regarding tempered distributions, your intuition is correct. They should produce real values on real-valued Schwartz functions $\varphi \in \mathcal{S}$. Its Fourier transform $\hat u$ would satisfy the same Hermitian symmetry $\hat u(y)=\overline{\hat{u}(-y)}$ (in an integrated sense) and so the same argument as above would give you a real inner product.
This is not so surprising because a function $u \in L^1_{\mathrm{loc}}$ is real-valued if and only if $\langle u,\varphi\rangle\in \mathbb{R}$ for all real-valued $\varphi \in C_c^\infty(\mathbb{R})$. So the fact that this is the natural generalisation to distributions makes sense.
On a side note, for more information on fractional Sobolev spaces and the various equivalent ways of defining them I would recommend the excellent review paper by Di Nezza, Palatucci, and Valdinoci: https://arxiv.org/pdf/1104.4345.pdf.