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Let $T$ be the operator from $L^2(\mathbb R^n)$ to $L^2(\mathbb R^n)$ that is given by $Tf := f * g$ where $g$ is in $L^2$.

How do I now find that the spectrum of $T$ is equal to the essential range of $\hat{g}$? How is the spectrum of $T$ related to the invertibility of the operator $G\hat{f} = \hat{f}\hat{g}$?

The hat denotes the Fourier transform.

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    It is interesting if for some function $a$ we consider the operator $Tf:= (af)*g$. Can we still find the spectrum of this operator?2011-01-29

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The point is that the Fourier transform $\mathcal{F}$ conjugates $T$ to the multiplication operator $S:h\mapsto\hat g h$ (i.e. $\mathcal{F}(f\ast g)=\hat g\mathcal{F}(f)$). Hence $T$ and $S$ have the same spectrum.

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    Then you use that $UAU^*$ and $A$ have the same spectrum. Okay.2010-09-18
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    Jonas, does that work in an infinite dimensional space?2011-01-29