Constructing a function $u$ such that $\int_{\mathbb{R}^2}|\eta-\xi||\hat{u}(\eta)|^2|\hat{u}(\xi)|^2\,d\xi\,d\eta<\infty$, but $u\notin H^{1/2}$

Question

For $u\in \mathcal{S}'(\mathbb{R})$, define, if finite, $$\Lambda(u)^4=\int_{\mathbb{R}^2}|\eta-\xi||\hat{u}(\eta)|^2|\hat{u}(\xi)|^2\,d\xi\,d\eta.$$ Using the triangle inequality $|\eta-\xi|\le |\eta|+ |\xi|$, we see that $$\Lambda(u)^4 \le 2 \| u\|_{L^2(\mathbb{R})}^2 \|u\|_{\dot{H}^{1/2}(\mathbb{R})}^2 \le 2\|u\|_{H^{1/2}(\mathbb{R})}^4.$$ Hence, if $X:=\{u\in \mathcal{S}'(\mathbb{R}): \Lambda(u)<\infty\}$, then $H^{1/2}(\mathbb{R})\subseteq X$.

My question is: how to construct $u$ such that $u\in X$ but $u\notin H^{1/2}(\mathbb{R})$, if that is possible.

Motivation: if $v$ solves the linear Schrödinger equation, \begin{aligned} \begin{cases} i\partial_t v+\Delta v =0 \\ v(0,x)=v_0 \end{cases} \end{aligned} then it can be shown remarkably that $$\|\partial_x(|v|^2) \|_{L^2_{t,x}(\mathbb{R}\times\mathbb{R})}^2=c \Lambda(v_0)^4$$ for some constant $c>0$. To see this, take the spacetime Fourier transform of $v\,\overline{v}$ and use Plancherel.

My first attempt was to try to find $u\in L^2\setminus H^{1/2}$ such that $|\eta-\xi|\ll \max(|\eta|, |\xi|)$ on the support of $|\hat{u}(\eta)|^2|\hat{u}(\xi)|^2$, which is $\operatorname{supp}\hat{u}_0 \times \operatorname{supp}\hat{u}_0$. For example, if we could use that $|\eta-\xi|\le 1$ on the support of $|\hat{u}(\eta)|^2|\hat{u}(\xi)|^2$, then $\Lambda(u)\le \| u\|_{L^2(\mathbb{R})}<\infty$, but there is no way to write the diagonal strip $\{(\eta, \xi): |\eta-\xi|\le 1\}$ as the square of a one-dimensional set.

I will also note that $\Lambda(u)$ looks like a multilinear pseudo-differential operator, so perhaps techniques from that field would be useful

Answer 1

(Un?)fortunately, there are no such functions. The idea is simple, yet powerful:

if the double integral $\int\int d\xi d\eta$ is finite, then for almost all $\eta$ the $d\xi$ integral is finite as well (Fubini--Tonelli or something). If the function $\hat{u}$ is zero almost everywhere then there is nothing to prove, everything is just zero. So, $\hat{u}$ is not zero on a set of positive measure. Let us fix two different $\eta_1\neq \eta_2$ for which $\hat{u}(\eta_1)\neq 0$ and $\hat{u}(\eta_2)\neq 0$ and for which the integral is finite. That is,

$$\int |\eta_1 - \xi||\hat{u}(\eta_1)|^2 |\hat{u}(\xi)|^2 d\xi < \infty, \int |\eta_2 - \xi||\hat{u}(\eta_2)|^2 |\hat{u}(\xi)|^2d\xi < \infty.$$

Let us add these expressions. We get $$\int (|\eta_1 - \xi||\hat{u}(\eta_1)|^2 + |\eta_2 - \xi||\hat{u}(\eta_2)|^2) |\hat{u}(\xi)|^2 d\xi< \infty.$$

And now I will leave it to you to convince yourself using the triangle inequality that $$(|\eta_1 - \xi||\hat{u}(\eta_1)|^2 + |\eta_2 - \xi||\hat{u}(\eta_2)|^2) \ge c(1+|\xi|)$$

for some constant $c > 0$ (it must depend on $\eta_1, \eta_2$ and $|\hat{u}(\eta_1)|$, $|\hat{u}(\eta_2)|$ of course). Hence, $$\int (1+|\xi|) |\hat{u}(\xi)|^2d\xi = ||u||_{H^{1/2}}^2< \infty.$$

Note, however, that there is no way to get a uniform bound $||u||_{H^{1/2}} \le C \Lambda(u)$, yet there are (sadly?) still no contradictions in mathematics, because $\Lambda$ is not a norm.