Let $f$ be a bounded function on $\mathbb R^n$,and the Fourier transform of $f$ is supported in the ball $B(R,0)$. Prove that there exists $C_{\alpha,n}$ such that $\|\partial_\alpha f\|_{L^\infty}\leq C_{\alpha,n}R^{|\alpha|}\|f\|_{L^\infty}$
Hint: consider $f * \phi $, where $\phi$ is a Schwartz function whose Fourier function equals $1$ in $B(R,0)$ and vanishes out of $B(2R,0)$.
But I don’t know how to amplify left hand side at all. Could you help me? Thanks!
Let $\varphi\in \mathcal{S}(\mathbb R^n)$ be a fixed function such that $\hat\varphi=1$ on $B(1, 0)$ and $\operatorname{supp}\hat\varphi\subset B(2, 0)$. For each $R>0$, let $$\phi_R(x)=R^n\varphi(Rx), \qquad x\in\mathbb R^n,\tag{$*$}$$ then $\hat\phi_R(\xi)=\hat\varphi(\xi/R)$, hence $\hat\phi_R=1$ on $B(R, 0)$ and $\operatorname{supp}\hat\phi\subset B(2R, 0)$. Note that $$\widehat{f*\phi_R}=\hat f\hat \phi_R=\hat f,$$ since $\hat\phi_R(\xi)=1$ if $\hat f(\xi)\neq0$. Therefore we have $f=f*\phi_R$, and thus $\partial_\alpha f=f*\partial_\alpha\phi_R$, so $$\|\partial_\alpha f\|_{L^\infty}\leq \|f\|_{L^\infty}\|\partial_\alpha\phi_R\|_{L^1}.$$
It follows from $(*)$ that $\partial_\alpha \phi_R(x)=R^{|\alpha|+n}(\partial_\alpha\varphi)(Rx)$, hence $$\|\partial_\alpha\phi_R\|_{L^1}=R^{|\alpha|}\|\partial_\alpha\varphi\|_{L^1}=C_{\alpha, n}R^{|\alpha|},$$ and this gives the desired inequality.