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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?

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Can you give an example how you change the inner product on $D$ such that $A$ becomes bounded? The implicit example of using the $H^1$ inner product on the first Sobolev space doesn't work as the momentum operator is still unbounded. –  Willie Wong Jan 27 '12 at 9:08
    
OK, let me write more precise what I mean. For a scalar product on $H^{1}$ I use $\langle f,g \rangle_{H^{1}} = \int (\overline{f(x)}g(x) + \overline{\frac{df}{dx}}\frac{dg}{dx})dx$ and the momentum operator is $P := -i\frac{d}{dx}$ which is a map $P:H^{1} \rightarrow L^{2}$. One easily sees that $\lVert Pf \rVert^{2}_{L^{2}} = \lVert f \rVert^{2}_{H^{1}} - \lVert f \rVert^{2}_{L^{2}} \leq \lVert f \rVert^{2}_{H^{1}}$. Therefore, for the norm of P we have that $\frac{\lVert Pf \rVert_{L^{2}}}{\lVert f \rVert_{H^{1}}} \leq 1$. Is that correct? –  physicits Jan 27 '12 at 11:11
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Uh... 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. –  Willie Wong Jan 27 '12 at 16:30

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