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Let $\phi \in \mathcal{S}(\mathbb R^d)$ and let $f:\mathbb R^d\to \mathbb C$ be function. Let $1\leq p \leq \infty.$ Question:

Can we say $\|\mathcal{F}^{-1} (\phi \hat{f})\|_{L^{p}} \leq C \|\hat{f}\|_{L^{p}}$? Can we say $\|\mathcal{F}^{-1} (\phi \hat{f})\|_{L^{p}} \leq C \|f\|_{L^{p}}$? $C$ is some constant.

(where $\mathcal{F}^{-1}$is inverse Fourier transform, and $\hat{.}$ is Fourier transform)

Motivation: What is a Bernstein's multiplier estimate? Is this applicable in this kind of situation?

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As Jose27 said, the second estimate is true by Young's convolution inequality because $\mathcal F^{-1}(\phi \hat f)$ is the convolution of $f$ with $\mathcal F^{-1}(\phi)$, and the latter is integrable.

The first inequality is true only for $p=2$. The Fourier transform is only bounded from $L^p$ to $L^q$ when $1\le p\le 2$ and $p^{-1}+q^{-1}=1$. Since you want $q=p$ here, $p=2$ is the only option. See Fourier transform in $L^p$.