By $L^p_T\dot H^s$ I mean the space $L^p([0, T] \to \dot H^s)$ where as usual $$\|u\|_{\dot H^s}^2 = \int_{\mathbb R^d} |\hat u(\xi)|^2 |\xi|^{2s} ~d\xi.$$ Of course, we can interpolate in the time norm $$\|u\|_{L^p_T\dot H^{2/p}} \leq \|u\|_{L^2_T \dot H^{2/p}}^{\frac{p}{2}} \|u\|_{L^\infty_T \dot H^{2/p}}^{\frac{2 - p}{2}}$$ and one may do the same in the space norm for each fixed time. However, I want to interpolate in both the time and space norms "at the same time." More precisely, I want something like $$\|u\|_{L^p_T\dot H^{2/p}} \leq \|u\|_{L^2_T \dot H^1}^{\frac{p}{2}} \|u\|_{L^\infty_T L^2}^{\frac{2 - p}{2}}$$ to be true. I know that we can extend the Riesz-Thorin theorem to mixed norms, but in that case (at least for the version of the theorem I know) the space norm varies in p, not in s, so that doesn't seem to help much here.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ To prove the inequalities you mention, you shouldn't need Riesz-Thorin (though I'm fairly sure complex interpolation works the way you want it to here). I think you can do those inequalities (with $L^q_t L^p_x$ spaces rather than $L^q_t H^s_x$ spaces) by hand with H{\"o}lder's inequality, and I think you can do them `by hand' with H{\"o}lder for $L^q_t H^s$ spaces as well. $\endgroup$– sharpendCommented Mar 29, 2021 at 3:17
-
1$\begingroup$ The exponents should be $2/p$ and $(p-2)/p$, by the way. Otherwise I agree with @sharpend: derive the inequality for $\dot{H}^{2/p}$ in terms of $\dot{H}^1$ and $L^2$ using the Hölder inequality and then just insert in the $L^p$ norm in time. $\endgroup$– HannesCommented Mar 29, 2021 at 8:47
-
$\begingroup$ Thank you both! I don't know how I managed to convince myself I needed something fancy like Riesz-Thorin, but following your comments I'm now convinced this follows from Holder's inequality. Etiquette question, would it be OK if I wrote up the details and submitted as an answer for future reference? $\endgroup$– Aidan BackusCommented Mar 29, 2021 at 18:04
Add a comment
|