Suppose $\frac{1}{p} + \frac{1}{q} =1 $ and $m \in L^{\infty} $, show that $m$ is $L^p$-multiplier if and only if $m$ is $L^q$-multiplier.

Question

My idea is to extend Hausdorff-Young inequality for $p > 2$, that is, suppose $f$ is a Schwartz function, then $\| \hat{f}\|_q \leq \| f\|_p$. By this, the completeness of $L^2$ and DCT:

$\| Tf\|_p = \| \check{m \hat{f}}\|_p \leq \| m\hat{f}\|_q \leq \| m\|_{\infty} \| \hat{f}\|_q \leq \| m\|_{\infty} \| f\|_p $

However, I don't know how to prove my conjecture.

Any hint?

Answer 1

Let $T$ be a bounded multiplier operator on $L^p.$ Consider the adjoint operator $T^*:L^q\to L^q.$ Then $T^*$ is bounded. Moreover for $f\in L^q $ and $g\in C^\infty_c$ we have $$\int (T^*f)g\,dx =\int f(Tg)\,dx=\int \hat{f}(m\hat{g})\,dx =\int (m\hat{f})\hat{g}\,dx $$ Therefore $T^*$ is the multiplier operator with the same symbol $m.$