Is there any example of linear operator which is bounded on all Besov spaces but not on Triebel-Lizorkin spaces

Turns out it was not known. So I took a week and write it down. You can see in the arXiv for a complete proof.

For convenience let $n=1$. Let $(\phi_j)_{j=0}^\infty$ be the family of Schwartz function to define Besov and Triebel-Lizorkin norm. More precisely $$\|f\|_{B_{pq}^s}:=\|(2^{js}\phi_j\ast f)_{j=0}^\infty\|_{\ell^q(L^p)};\quad\|f\|_{F_{pq}^s}:=\|(2^{js}\phi_j\ast f)_{j=0}^\infty\|_{L^p(\ell^q)}.$$ Special modification is required for $p=\infty$ but let's omit that here.

Take $(y_j)_{j=0}^\infty\subset\mathbb R$ to be a sequence of uniformly discrete points, e.g. $y_j=j$. The model operator I choose is $$Tf(x)=\sum_{j=0}^\infty\phi_j\ast f(x-y_j).$$

For Besov spaces the frequency information are separated therefore $T:B_{pq}^s\to B_{pq}^s$ is bounded for all $p,q,s$.

However $T:F_{pq}^s\not\to F_{pq}^s$ when $p\neq q$.

To see this we pick $f(x)=\sum_{j=0}^\infty 2^{-js}a_j\cdot\chi(x)e^{-2\pi i2^jx}$ where $\chi\in C_c^\infty(0,1)$ is fixed and $(a_j)_j\subset\mathbb R$.

Then roughly speaking $2^{js}\phi_j\ast f(x)\approx a_j\chi(x) e^{-2\pi i2^jx}$ for all $j$ large. Therefore $$\|f\|_{F_{pq}^s}\approx\|(a_j\chi)_j\|_{L^p(\ell^q)}\approx\|a\|_{\ell^q};\quad\|Tf\|_{F_{pq}^s}\approx\|T^{-1}f\|_{F_{pq}^s}\approx\|(a_j\chi(\cdot-y_j))_j\|_{L^p(\ell^q)}\approx\|a\|_{\ell^p}.$$

When $p<q$ take $a\in\ell^q\backslash\ell^p$ we get $f$ for a counterexample. When $p>q$ we take $a\in\ell^p\backslash\ell^q$, then $T^{-1}f$ will be a counterexample.