2
$\begingroup$

Suppose that $m:\mathbb R \to \mathbb C$ satisfies: there exists $C > 0$ such that

$$\| (m \hat{f})^{\vee} \|_{L^{p}} \leq C \|f\|_{L^{p}}.$$

That is, $m$ is an $L^{p}$-multiplier. Let $M(L^{p})$ denote the space of $L^p$-multipliers.

Put $m_{\delta}(x)= m(\delta x), (\delta >0),$ $T_{k}m(x)=m(x-k), (k\in \mathbb R).$

My questions:

(1) If $m\in M(L^p) $, then can we say that $m_\delta, T_{k}m \in M(L^{p})$?

(2) Let $m\in M(L^p)$ and $\phi : \mathbb R \to \mathbb R$ be such that $m\circ \phi \in M(L^{p})$. What can we say about $\phi$?

If this is known, references would be helpful.

$\endgroup$
1
  • $\begingroup$ Q1 has been answered below. For Q2 as phrase, the answer is literally nothing, since $m \equiv 0$ is a (trivial) $L^p$ multiplier. $\endgroup$ Dec 14, 2020 at 15:03

1 Answer 1

3
$\begingroup$

For $1 \leq p < \infty$, let $M( L^p(\mathbb{R}))$ be the space of all $m \in L^{\infty} (\mathbb{R})$ such that the operator $$ \mathcal{F}^{-1} (\hat{f} \cdot m), \quad f \in \mathcal{S} (\mathbb{R}),$$ with $\mathcal{F} : \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R})$ denoting the Fourier transform, is bounded on $L^p (\mathbb{R})$. The norm of $m \in M (L^p (\mathbb{R}))$ is defined as $$ \| m \|_{M(L^p (\mathbb{R}))} = \| T_m \|_{L^p \to L^p}, $$ and one can show that $(M(L^p (\mathbb{R})), \| \cdot \|_{M(L^p (\mathbb{R})})$ forms a Banach algebra.

Regarding your question, it is stated as Proposition 2.5.14 in Grafakos' Classical Fourier Analysis that $$ \| T_k m \|_{M(L^p (\mathbb{R}))} = \| m \|_{M(L^p (\mathbb{R}))}$$ and $$ \| M_{\delta} m \|_{M(L^p (\mathbb{R}))} = \| m \|_{M(L^p (\mathbb{R}))}, $$ whose proof is left as an exercise. However, I think this follows immediately from the fact that $T_k$ and $M_{\delta}$ are isometric operators on $L^p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.