I tried to find this example but the condition $\overline{\operatorname{range}(\lambda I -A)}=C[0,1]$ is too hard to prove. Anyone could help me?
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Consider the Hilbert space $H=L^2((0,1))$ and let $A:D(A) \to H$ be defined by
$(Af)(x)=xf(x),$ where $D(A)=\{f \in H: xf \in H\}.$
Then $A$ is self-adjoint, hence the residual spectrum of $A$ is empty. Furthermore, $A$ has no eigenvalues.
Conclusion:
$$\sigma(A)= \sigma_c(A)=[0,1].$$
Fred
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$D(A)=H$....... – daw May 24 '19 at 08:42
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1This is a good example, but I need to work in C[0,1] space, which is not a Hilbert Space – Aquila May 24 '19 at 13:27
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@daw is correct as $$| x \mapsto x \cdot f(x) |_2^2 = \int_0^1 x^2 | f(x) |^2 dx \le \int_0^1 | f(x) |^2 dx = | f |_2^2 < \infty$$ – ViktorStein Mar 10 '20 at 21:54