I have come across various instances of interpolation estimates of the following type:


$ \| f \|\_{L^r(\mathbb{R^n})} \lesssim \| |\nabla|^{-s\_1} f \|\_{L^{r\_1}(\mathbb{R^n})}^\theta \| |\nabla|^{s\_2} f \|\_{L^{r\_2}(\mathbb{R^n})}^{1-\theta}$,


where


$\tfrac{1}{r} = \tfrac{\theta}{r\_1}+\tfrac{1-\theta}{r\_2}$


and $s\_1,s\_2>0$ satisfy


$-s\_1\theta + s\_2(1-\theta) =0$.


Here $|\nabla|^s$ are the usual fractional derivative operators defined via the Fourier transform. I am not sure what restrictions on the exponents are needed beyond $1\leq r,r\_1,r\_2\leq \infty$ (perhaps excluding the endpoints).


I was hoping that someone could help me understand what type of interpolation is being used here and precisely how it is being applied. 


In fact, I believe that one can prove this estimate directly for $1<r,r\_1,r\_2<\infty$ via the square function estimate and Hölder's inequality (first in the sums and then in the integrals), using 


$|P\_N f|^2 = ( N^{-2s\_1}|P\_Nf|^2)^\theta (N^{2s\_2}|P\_N f|^2)^{1-\theta}$


for an individual Littlewood-Paley piece.


But, clearly the estimate should follow from some appropriate interpolation, and I would be very grateful to understand exactly what type of interpolation is being used.



An interpolation estimate involving fractional derivatives
I would add this as a comment, but I dont have enough reputation.


If I understand you correctly, then your $|| |\nabla|^{-s\_1}||\_{L^{r\_1}(\mathbb{R}^n)}$ is just the norm of the Bessel potential space $H^{-s\_1}\_{r\_1}(\mathbb{R}^n)$. For the complex interpolation method we have (using your notation)


$(H^{s\_2}\_{r\_2}(\mathbb{R}^n), H^{-s\_1}\_{r\_1}(\mathbb{R}^n))\_{[\theta]}=H^0\_r(\mathbb{R}^n)$


for $0<\theta<1$ and $1<r\_1,r\_2<\infty$.


Is it this kind of result you were looking for?


