There is a question that puzzles me, so may be someone here has an answer. Assume we have a symmetric operator $A$ that is defined on a space $D$ that is dense in $L^2$, so $A:D\rightarrow L^2$, and $A$ is unbounded when one uses the usual $L^2$ inner product. For example, the momentum operator defined on the first Sobolev space. Assume we know the spectrum of $A$ in this case and it consists of the entire real line. Now, we change the inner product on $D$ such that $D$ is becomes a Hilbert space and $A$ becomes bounded. Therefore, also the spectrum of $A$ should now be bounded. How can one understand this fact and relate the two spectra to each other through the two scalar products?
relation between inner product and spectrum
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spectral-theory
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1Uh... if $P:H^1 \to L^2$, what do you mean by the spectrum of $P$? Usually when you study the spectrum of a (densely defined) operator, you use the same Banach space on both sides, I'm not familiar with this attempt at embedding $Dom(P)$ and $Ran(P)$ into distinct normed spaces. – 2012-01-27