# $L^\infty$ bound on smooth compactly supported function using Fourier transform

Let $u \in C^\infty_c(\mathbb R^d)$ and $s> \frac{d}{2}$. Show that

$$||u||_{L^\infty}^2 \leq K \int_\mathbb{R}^d |\widehat{u}(\zeta)|^2(1+|\zeta|)^{2s} d\zeta,$$ for some constant $K=K(d,s)$.

Thoughts: Not sure where to start. The $(1+|\zeta|)^{2s}$ makes me think of the rate of decay of the Fourier transform given that its smooth and of compact support. Other than that, not really sure. Don't see how to apply any of the classic inequalities here either.

• Hmm so in the C-S inequality something like $||\frac{1}{(1+|\zeta|)^{2s}}||_2^2 ||\widehat{u}(\zeta) (1+|\zeta|)^{2s}||_2^2$ on the RHS? Oct 6 '16 at 3:39
• Basically, but no $2s$ because there is already a square from the $L^2$-norm. Oct 6 '16 at 3:53