Let $\delta_n\in C^0_c(\mathbb{R})$ be a Dirac sequence approximating the Dirac delta "function" $\delta$ with support in $0\in \mathbb{R}$. Then, for each $n$ we have a compact operator $K_n:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ given by multiplication with $\delta_n$. My question is: How changes the spectrum of the compact opearator $K_n$ as $n\rightarrow \infty$ ?
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3$\begingroup$ The multiplication operator is not compact, and its spectrum is the (essential) range of the function you multiply with. $\endgroup$– Christian RemlingFeb 24, 2019 at 9:13
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$\begingroup$ @ChristianRemling: I think what the OP actually means is that $K_n$ the convolution with $\delta_n$ (so the spectrum of $K_n$ is the range of the Fourier transform of $\delta_n$). $\endgroup$– Jochen GlueckFeb 24, 2019 at 11:46
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$\begingroup$ @JochenGlueck: Ok, but then we're still dealing with a multiplication operator after taking FTs, and my comments still apply. $\endgroup$– Christian RemlingFeb 24, 2019 at 17:32
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$\begingroup$ @ChristianRemling: Fair enough :-). $\endgroup$– Jochen GlueckFeb 24, 2019 at 19:40
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