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Is the transformation $f\mapsto T(f):=\mathcal{F}^{-1}(m(\xi)\widehat{f}(\xi))$ always translation-invariant?

My attempt: (with $f$ a Schwartz function) \begin{align} (T (\tau_hf))(x)&=\mathcal{F}^{-1}(m(\xi)\widehat{\tau_hf}(\xi))(x))\\ &=\int_{\mathbb{R}}\mathrm{e}^{ix\xi}m(\xi)\widehat{\tau_h f}(\xi)\,d\xi\\ &=\int_{\mathbb{R}}\mathrm{e}^{ix\xi}m(\xi)\mathrm{e}^{ih\xi}\widehat{f}(\xi)\,d\xi\\ \end{align} and \begin{align} (\tau_h (Tf))(x)&=\tau_h(\mathcal{F}^{-1}(m(\xi)\widehat{f}(\xi))(x)\\ &=\mathcal{F}^{-1}(m(\xi)\widehat{f}(\xi))(x+h)\\ &=\int_{\mathbb{R}}\mathrm{e}^{i(x+h)\xi}m(\xi)\widehat{f}(\xi)\,d\xi\\ &=\int_{\mathbb{R}}\mathrm{e}^{ix\xi}m(\xi)\mathrm{e}^{ih\xi}\widehat{f}(\xi)\,d\xi \end{align}

Therefore, $\tau_h T=T\tau_h$?, for any $m(\xi)\in L^{\infty}(\mathbb{R}))$ Or does the function $m(\xi)$ need some additional assumptions?

Feng
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eraldcoil
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  • this is called commute with translation. and yes, because the translation is multiplication operation. – Yimin Dec 27 '23 at 04:45
  • possible duplicate https://math.stackexchange.com/questions/1354210/operator-t-commutes-with-all-translations-leftrightarrow-t-is-multiplicat?rq=1 – Yimin Dec 27 '23 at 04:46
  • @Yimin. But, it's sufficient that $m(\xi)$ is bounded? Where I can find a reference? thanks! – eraldcoil Dec 27 '23 at 04:50
  • @eraldcoil deps on what space your T defined with, say Schwartz to $L^\infty$. Then it should be correct. – Yimin Dec 27 '23 at 05:01

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