Given a positive integer $n$ and a>0. Let consider the operators $\nabla^n (\cdot)= \sum_{i=1}^d \partial_i^{n}(\cdot) $, and $(1- \Delta)^{\frac a 2} $ defined at the Fourier level as (modulus some numerical constant) $$ \widehat{(1- \Delta)^{\frac a 2} } (\xi) = (1- |\xi|^2)^{\frac a 2}. $$
I am wondering if the following inequality holds:
$$ \langle \ \nabla^n (1- \Delta)^{\frac a 2} f , \nabla^n (1- \Delta)^{\frac a 2} g \ \rangle_{L^2(\mathbb{R}^d)} \leq C \langle \ (1- \Delta)^{\frac {a+n} 2} f , (1- \Delta)^{\frac {a+n} 2} g \ \rangle_{L^2(\mathbb{R}^d)}. $$
For me it follows from considering Plancherel Theorem and $$ |\xi|^n = |\xi|^{2 \frac n 2} \leq (1+ |\xi|^2)^{\frac n 2}, $$
but I'm not sure of it. Do you have any comment or idea?
In fact my strategy is the following:
By Plancherel we have $$ \langle \ \nabla^n (1- \Delta)^{\frac a 2} f , \nabla^n (1- \Delta)^{\frac a 2} g \ \rangle_{L^2(\mathbb{R}^d)} = \langle \ (i)^n |\xi|^n (1+|\xi|^2)^{\frac a 2} \widehat{f} , (i)^n |\xi|^n (1+|\xi|^2)^{\frac a 2} \widehat{g}\ \rangle_{L^2(\mathbb{R}^d)} $$
and considering $|\xi|^n \leq (1+ |\xi|^2)^{\frac n 2},$
$$ \langle \ (i)^n |\xi|^n (1+|\xi|^2)^{\frac a 2} \widehat{f} , (i)^n |\xi|^n (1+|\xi|^2)^{\frac a 2} \widehat{g}\ \rangle_{L^2(\mathbb{R}^d)} \leq \langle \ (i)^n (1+|\xi|^2)^n (1+|\xi|^2)^{\frac a 2} \widehat{f} , (i)^n (1+|\xi|^2)^n (1+|\xi|^2)^{\frac a 2} \widehat{g}\ \rangle_{L^2(\mathbb{R}^d)}. $$