# How fourier transform of a tempered distribution could be a function (Homogeneous Sobolev space)?

I met the following definition of homogeneous Sobolev space: $$\dot{H^s}(\mathbb{R}^n) = \left\{ f \in \mathcal{S}': \hat{f} \in L_{loc}^1(\mathbb{R}^n) \,\,\, \text{and} \,\,\, || f||_{\dot{H^s}} < \infty \right\}$$,
where $$s \geq 0$$ (does not have to be an integar), $$\hat{f}$$ is the fourier transform of $$f$$,
$$|| f||_{\dot{H^s}} =\left(\displaystyle\int_{\mathbb{R}^n}^{} |\xi|^{2s} |\hat{f}(\xi)|^2 \, d\xi \right)^\frac{1}{2}$$ and
$$\mathcal{S}'$$ is the space of linear and continuous functionals from Schwarz Space $$\mathcal{S}$$ to $$\mathbb{C}$$ (called space of tempered distribiutions).
If $$F \in \mathcal{S}'$$ and $$f \in \mathcal{S}$$, then the fourier transform on it can be defined as $$\hat{F}(f) = F(\hat{f}).$$
And my question is: if $$g \in \dot{H^s}(\mathbb{R}^n)$$ then how $$\hat{g} \in L_{loc}^1(\mathbb{R}^n)$$, I mean how $$\hat{g}$$ can be a function if $$g$$ is a functional?
Similarly what does mean the integral defined in $$||.||_{\dot{H^s}}$$ provided that i do not know if $$\hat{g}$$ is a function?

For simplicity let me assume that $$s$$ is a positive integer.

You should interpret $$\int_{\mathbb{R}^n}^{} |\xi|^{2s} |\hat{f}(\xi)|^2 \, d\xi<\infty$$ as meaning that $$f$$ is a tempered distribution with Fourier transform $$\hat{f}$$ so that $$(\sum_{j=1}^n \xi_j^2)^s \hat{f}$$ is a tempered distribution which is represented by a function $$g\in L^2(\Bbb{R}^n)$$, that is $$\langle \hat{f},\phi \rangle= \int_{\Bbb{R}^n} g(\xi)\phi(\xi)d\xi$$ for all $$\phi \in S(\Bbb{R}^n)$$.

So it is allowed to say that $$|\xi|^{2s} \hat{f} \in L^2(\Bbb{R^n})\subset L^1_{loc}(\Bbb{R^n})$$.

It doesn't have to be that $$\hat{f}$$ is $$L^1_{loc}$$.

What you can say is that taking $$\psi\in C^\infty_c(\Bbb{R}^n), \psi=1$$ on $$|\xi|<1$$ then $$T(\phi)=|\xi|^{-2s}(\phi-\psi \sum_{|\alpha|<2s} \frac{\xi^\alpha}{\alpha!}\partial^\alpha \phi(0))$$ defines a continuous map $$\phi \to T(\phi), S(\Bbb{R}^n)\to S(\Bbb{R}^n)$$, and the tempered distribution $$\langle h,\phi \rangle = \langle \hat{f},\phi \rangle - \langle g,T(\phi) \rangle$$

is of order $$\le 2s$$ and supported at $$0$$. So that $$h = \sum_{|\alpha|\le 2m} c_\alpha \partial^\alpha \delta$$ and $$\langle \hat{f},\phi \rangle = \sum_{|\alpha|\le 2m} c_\alpha \partial^\alpha \phi(0) + \int_{\Bbb{R}^n} g(\xi) |\xi|^{-2s}(\phi(\xi)-\psi(\xi) \sum_{|\alpha|<2s} \frac{\xi^\alpha}{\alpha!}\partial^\alpha \phi(0)) d\xi$$

• So, if I understand properly, the condition $\hat{f} \in L_{loc}^1$ means $\hat{f}$ is represented by some $g \in L_{loc}^1?$ There is also a theorem saying that for a very specific s and p $\dot{H^s}(\mathbb{R}^n) \subset L^p(\mathbb{R}^n)$. Does it just say that $h \in \dot{H^s}$ has a representative in $L^p$? Commented Mar 12, 2023 at 10:02
• I also wonder if expressions like for example $\displaystyle\int_{A} e^{2 \pi i x \xi}\hat{f}(\xi)\ d\xi$ could be somehow sensible for $f \in \dot{H^s}$ and some sensible set $A$, because in some works I saw people writing similar things and i am still not sure what "$\hat{f}(\xi)$" means. Commented Mar 12, 2023 at 10:19