$\int_{|x|<t} |\mathcal{F}^{-1}f(x) |dx \leq C t^{\frac{n}{2}}\|f\|_{L^{2}}$?

Question

Let $f\in L^{2}(\mathbb R^{n}).$

Fix $t>0,$

My Question:How to show, $\int_{|x|<t} |\mathcal{F}^{-1}f(x)| dx \leq C t^{\frac{n}{2}}\|f\|_{L^{2}}$ ?

[We note $\mathcal{F}$ denotes the Fourier transform and $\mathcal{F}^{-1}$ the inverse Fourier transform; and $C$ is a constant]

My Motivation: This is one of the key step in proving Bernstein multiplier theorem.

Thanks,

Answer 1

1/The Fourier transform can be viewed as a unitary operator on $L^2(\mathbb{R}^n)$, which means that if $f \in L^2(\mathbb{R}^n)$, then $\mathcal{F}[f]$ and $\mathcal{F}^{-1}[f]$ too.

2/We have from the Parseval equation: $\frac{1}{2 \pi} \int_\mathbb{R} \vert f(x) \vert ^2 dx = \int_\mathbb{R} \vert \mathcal{F}[f](x) \vert ^2 dx$

3/Now $\int_{|x|<t} |\mathcal{F}^{-1}f(x)| dx = \int_\mathbb{R^n} 1_{| x| <t } |\mathcal{F}^{-1}f(x)| dx$ is the integral of two functions in $L^2(\mathbb{R}^n)$. Applying the Cauchy-Schwarz inequality, we have : $\int_{|x|<t} |\mathcal{F}^{-1}f(x)| dx \leq \sqrt{\int_\mathbb{R^n} 1_{| x| <t }dx} \sqrt{\int_\mathbb{R^n} |\mathcal{F}^{-1}f(x)|^2dx} $