Construct a function exactly belongs to $H^1(\mathbb{R})$

Question

Question: How to construct a function exactly belongs to $H^1(\mathbb{R})$ but does not belong to $H^{1+s}(\mathbb{R})$ for any $s>0$?

My try: An obviously try is $u(x)=|x|$. However, I find that $u(x)\in H^{1+1/2-\epsilon}(\mathbb{R})$ actually. Is there any way to find out this function? Thanks in advance.

Answer 1

Something like $$ u(x) := \int_0^x \frac{e^{-t^2}}{\sqrt{|t|} \log |t|} \, dt $$ should do the work.

Answer 2

Let $f$ be the inverse Fourier transform of $$\hat{f}(\xi) = \frac{\sqrt{2|\xi|}}{(1+\xi^2)\log(2 + 2\xi^2)}$$ Then $$(1+ \xi^2)|\hat{f}(\xi)|^2 = \frac{2|\xi|}{(1+\xi^2)\log^2(2+2\xi^2)} = - \text{sgn}(\xi)\frac{d}{d\xi}\left(\frac{1}{\log(2+2\xi^2)}\right)$$ It follows that $$\int_{\mathbb{R}}(1+ \xi^2)|\hat{f}(\xi)|^2 d\xi = \frac{2}{\log(2)}$$ and for $s>0$ $$\int_{\mathbb{R}}(1+ \xi^{2+2s})|\hat{f}(\xi)|^2 d\xi = +\infty$$ Hence $f\in H^1(\mathbb{R})$ but $f\notin H^{1+s}(\mathbb{R})$.

Edit As suggested by @TZakrevskiy in the comment, the same method gives a function that belongs to $H^s(\mathbb{R})$ but not to $H^{s+\epsilon}(\mathbb{R})$ for any $\epsilon>0$. One can choose $$\hat{f}(\xi) = \sqrt{\frac{2|\xi|}{(1+\xi^{2s})(1+\xi^2)}}\frac{1}{\log(2 + 2\xi^2)}$$