Let $u \in L^2(\mathbb{R}^n)$ be a measurable function such that $\Delta u - (-\Delta)^{\lambda} u \in L^2(\mathbb{R}^n)$ for $\lambda \in (0,1)$. Does that imply each of $\Delta u$ and $(-\Delta)^{\lambda} u$ is in $L^2(\mathbb{R}^n)$? Here $(-\Delta)^{\lambda}$ denotes the fractional laplacian.
I considered Fourier transform, $\mathcal{F}({\Delta u - (-\Delta)^{\lambda} u})= -|\xi|^2\hat u(\xi)-|\xi|^{2\lambda} \hat u(\xi)$ and we know $u \in L^2(\mathbb{R}^n)$ implies $\hat u \in L^2(\mathbb{R}^n)$. I'm not sure how to conclude using these informations.
$\newcommand{\bbR}{\mathbb{R}}$Certainly, it is enough to prove that $(-\Delta)^\lambda u$ is in $L^2(\bbR^n)$. Moreover, since the Fourier transform and its inverse are bounded on $L^2$, it suffices to prove that $|\xi|^{2\lambda}\hat{u} \in L^2$.
No this end, notice that $$|\xi|^{2\lambda} \hat{u} = \left\{\frac{|\xi|^{2\lambda}}{|\xi|^2 + |\xi|^{2\lambda}}\right\}\left(|\xi|^2 + |\xi|^{2\lambda}\right) \hat{u}.$$ The term in curly braces is in $L^\infty$, while the second term is in $L^2$ by assumption, so we can apply Holder's inequaltiy to conclude that $(-\Delta)^\lambda u \in L^2$.