We are attempting to show that, for $t_0,t_1\in\mathbb{R}$, and $t_\sigma=(1-\sigma)t_0+\sigma t_1$ for any $\sigma\in[0,1]$, then the following inequality holds for any $u\in H^{t_1}(\mathbb{R}^d)$:
$\|u\|_{H^{t_\sigma}(\mathbb{R}^d)}\leq\|u\|^{1-\sigma}_{H^{t_0}(\mathbb{R}^d)}\cdot \|u\|^\sigma_{H^{t_1}(\mathbb{R}^d)}$
A hint given to us was to write the LHS as a Fourier transform, but we do not see how to go about doing that nor how that would help. Any assistance in this area would be greatly appreciated. Thank you.
We have, by definition of $H^{t_\sigma}(\def\R{\mathbf R}\R^d)$, that $$ \def\norm#1{\left\|#1\right\|}\def\F{\mathscr F} \norm{u}_{H^{t_\sigma}} = \norm{ (1 + \def\abs#1{\left|#1\right|}\abs\cdot^2)^{t_\sigma/2}\F u}_{L^2} $$ We have for $\frac 1\alpha + \frac 1\beta = \frac 12$ by Hölder \begin{align*} \norm{u}_{H^{t_\sigma}} &= \norm{ (1 + \def\abs#1{\left|#1\right|}\abs\cdot^2)^{t_\sigma/2}\F u}_{L^2}\\ &= \norm{\def\f{(1+\abs\cdot^2)}\abs{\f^{t_0/2}}^{1-\sigma}\abs{\F u}^{1-\sigma} \cdot \abs{\f^{t_1/2}}^\sigma\abs{\F u}^\sigma}_{L^2}\\ &\le \norm{\abs{\f^{t_0/2} \F u}^{1-\sigma}}_\alpha \norm{\abs{\f^{t_1/2}\F u}^{\sigma}}_\beta \end{align*} We want the first term to be $\norm{u}_{H^{t_0}}$, that is, $\alpha = \frac{2}{1-\sigma}$, then $$ \beta = \frac 1{\frac 12 - \frac 1\alpha} = \frac 1{\frac \sigma 2} = \frac 2\sigma $$ so we have \begin{align*} \norm{u}_{H^{t_\sigma}} &\le \norm{\abs{\f^{t_0/2} \F u}^{1-\sigma}}_\alpha \norm{\abs{\f^{t_1/2}\F u}^{\sigma}}_\beta\\ &= \norm{\f^{t_0/2} \F u}_{L^2}^{1-\sigma} \norm{\f^{t_1/2}\F u}_{L^2}^{\sigma}\\ &= \norm{u}_{H^{t_0}}^{1-\sigma}\norm{u}_{H^{t_1}}^\sigma \end{align*}