1.$$ Hence $T_p(b_1,b_2)<\infty.$

On the other hand by $-2q/(p+q) < -1,$ the singularity in $\int_0^1{|\psi-\langle\psi\rangle_{[0,1)}|}^q$ is not integrable, and by $\int_0^1{|\phi-\langle\phi\rangle_{[0,1)}|}^q > 0,$ we have $S_q(\psi,\phi)=\infty.$

If one wishes to have $\psi,\phi\in L^{\infty}_{loc},$ say to have the joint conditions well-defined, Proposition can be modified by considering multiple copies of the situation spread out through ${\mathbb{R}}$ and introducing the singularities in $\psi$’s only gradually as is done in the next theorem.

\[nip1\] There exist functions $\psi,\phi \in L^{\infty}_{loc}$ failing the condition $S_{2+\varepsilon}$ for all $\varepsilon > 0,$ such that $[\phi,[\psi,H]]:L^2\to L^2$ boundedly, where $H$ is the Hilbert transform, i.e. the 1-dimensional Riesz transform.

By Theorem \[lowerbound\] the $L^2$ boundedness implies that $\psi,\phi$ satisfy the conditions $T_2,S_2.$

Let $$\psi_0^k(x) = c_kx^{-\eta_k}1_{(c_k^{6k}e^{-100k^2},1)}(x), \quad \eta_k = \frac{1}{2+k^{-1}},\qquad
        \phi_0(x) = x^{1/2}1_{(0,1)}(x),$$ where $c_k$ depends on $k$ and will be determined later. Let $\tau_hf(x) = f(x-h).$ Then set $\phi_k = \tau_k\phi_0$ and $\psi_k = \tau_k\psi^{k}_0$. Finally we define $$\phi = \sum_{k\in 2\mathbb Z }\phi_k,\qquad \psi = \sum_{k\in 4\mathbb N +2}\psi_k.$$ Let $k\in 4\mathbb N +2$ be fixed. We first show that the pair $(\psi_k, \phi)$ satisfies $S_q,T_q$ for $q=q_k=2+k^{-1}/2.$ Since $\tau_{k}\phi=\phi$ it suffices to prove that $(\psi_0^k, \phi)$ satisfies $S_q,T_q$. Again, for any interval $I$, we 