# Fourier multipliers on $L^2(\mu)$

On $$L^2(\mathbb{R}^d)$$, we have $$T_m$$ defined $$\widehat{T_m f} = m \widehat{f}$$ is a bounded operator on $$L^2$$ if and only if $$m \in L^\infty$$.

What can be said about the same problem for more general measures? For example, I am interested in weighted $$L^2$$ spaces where $$\mathrm{d}\mu(x) = (1 + |x|^{2})^{s/2} \mathrm{d}x,$$ when is $$T_m$$ a bounded operator on $$L^2(\mu)$$?

• How did you show the first statement? If I use that $\mathcal F$ is unitary (Plancherel) in $(L^2, \|.\|_2)$, and then use Cauchy-Schwartz, I get that the operator norm is $\|m\|_{2}$. Commented Oct 4, 2023 at 7:11
• We have $\| T_mf \|_2 = \big( \int m(\xi)^2 \widehat{f}(\xi)^2 \mathrm{d}\xi \big)^{1/2} \leq \|m\|_\infty \| \widehat{f} \|_2 = \|m\|_{\infty} \|f\|_2$
– Jack
Commented Oct 6, 2023 at 9:26

Let's assume $$s>0$$. Then $$\mathcal{F} (L^2(\mu))$$ is the Sobolev space $$H^{2,s}$$ (functions with $$s$$ derivatives in unweighted $$L_2$$), and your weighted space norm is giving the Sobolev norm of $$\widehat{f}$$.
To be bounded, $$T$$ can't destroy the smoothness of $$\widehat{f}$$, so you would want $$m$$ to be in some space like the $$L_\infty$$ Sobolev space of the same smoothness.