Does there exist a self-adjoint operator whose spectrum is just the continuous spectrum, i.e. no point spectrum and no residue spectrum?
If not, please prove it.
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6$(Mf)(x)=xf(x)$ on $L^{2}[0,1]$ has continuous spectrum $[0,1]$ and no other type of spectrum. – Disintegrating By Parts Jan 12 '16 at 10:16
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2And self-adjoint operators do not have residual spectrum anyway. – MaoWao Jan 12 '16 at 10:18
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As a note to @DisintegratingByParts comment/answer, you can see that $\sigma(M) = \sigma_c(M) = [0,1]$, in this answer of his. – ViktorStein Mar 10 '20 at 21:33
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2Does this answer your question? I want to find an example where the Spectrum is equal to the Continuous Spectrum in C[0,1]; $\sigma_c(A)=\sigma(A)$ – ViktorStein Mar 10 '20 at 21:40