It is known that $\vert x\vert^{-1}$ is bounded from $H^{1}(\mathbb{R}^3)$ in $L^{2}(\mathbb{R}^3)$ and this can be shown using, for example, Hardy's inequality. Said that, I would you like someone help me to calrify the following statement. On my notes, I have written that this implies that $\vert x\vert^{\frac{1}{2}}$ is bounded from $H^{\frac{1}{2}}(\mathbb{R}^3)$ by interpolation.
What does "by interpolation" mean? And how to show that $\vert x\vert^{\frac{1}{2}}$ is bounded from $H^{\frac{1}{2}}(\mathbb{R}^3)$?
Could anyone help me? Thank You in advance!
I suppose it should be $|x|^{-1/2}$ bounded from $H^{1/2}$ to $L^2$?
Yes, this is a result of interpolation theory. It is a theory that allows to find properties for intermediate spaces. There are two main settings: real and complex interpolation. You should look at https://en.wikipedia.org/wiki/Interpolation_space, or for example the book by Luc Tartar about "An Introduction to Sobolev Spaces and Interpolation Spaces" if you want to know more about it.
In your case, $H^{1/2}$ is the complex interpolation of $H^1$ and $L^2$ while the complex interpolation space between $L^2$ and $L^2$ with a weight $|x|^{-1}$ is the space $L^2$ with a weight $|x|^{-1/2}$. And you deduce the result by saying that the identity map is bounded between the two pair of spaces, and so also for the intermediate pair.