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Fix $l^{2}$ space and suppose we have the right shift operator and left shift operator. The goal is to find the point spectrum, continuous spectrum and residual spectrum of both operators using basic definitions.

This question has been asked frequently on the site, such as
Spectrum of shift-operator but most don't address my needs.

The point spectrum of both operators are easy to determine. For right-shift operator $R$, the kernel of $R - \lambda I$ would always be $0$ regardless of $\lambda$. On the other hand, for left-shift operators one can always find $z=(1,\lambda, \lambda^{2}, ...)$ so that any $|\lambda| \lt 1$ would be a point spectrum (the case when $\lambda = 0$ is also easy).

But what happens when $\lambda = 1$? How come, for example, the sequence $(x_2 - x_1, x_3 - x_2, ...)$ will be dense in $l^{2}$ but $(-\lambda x_1, x_1 - \lambda x_2, ...)$ not dense when $|\lambda| < 1$? My idea is to explicitly find a nonzero orthogonal vector to show the range not dense...

Dinoman
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  • The range of an operator $T$ is dense iff $T^$ is injective. Apply this to $S-\lambda I,$ where $S$ is the shift operator. Since you have identified the point spectrum of $S$ and $S^,$ the problem becomes easy. By the way the point spectrum of the left shift corresponds to $|\lambda|<1.$ – Ryszard Szwarc May 16 '24 at 01:51

1 Answers1

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If $|\lambda|<1$ then the vector $(1,\lambda,\lambda^2,\ldots)$ is orthogonal to the range of the operator $(-\lambda x_1, x_1-\lambda x_2,\ldots).$ You can verify this by direct computation of the scalar product.

Lieven
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