I am reading the book Index Theorem and the Heat Equation written by Peter.B.Gilkey. Here is my question: Let E be a hermitian vector bundle on a compact smooth manifold M. Let $D : \mathcal{C}^{\infty}(E) \rightarrow \mathcal{C}^{\infty}(E)$ be an elliptic operator of positive order. Then it extends to the completion of $\mathcal {C}^{\infty}(E)$ to $L^2(E)$. It has been said in the first paragraph of chapter 3 of the book mentioned above that this is an unbounded operator and admits a sequence of eigenvalues $0 \leq \lambda_n \rightarrow \infty$. Any reference or proof of this will be helpful.
closed as offtopic by Chris Gerig, Michael Renardy, Johannes Ebert, Stefan Kohl, Yemon Choi Sep 8 '14 at 15:08
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 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Chris Gerig, Michael Renardy, Johannes Ebert, Stefan Kohl

$\begingroup$ This follows from the spectral theorem for self adjoint unbounded operators together with elliptic regulatory: $L^2(E)$ decomposes as the direct sum of finite dimensional eigenspaces for $D$. Any book on PDE's will prove this. $\endgroup$ – Paul Siegel Sep 8 '14 at 13:06

$\begingroup$ Also could you please point out why the operator will be unbounded. Thanks. $\endgroup$ – ankit Sep 8 '14 at 13:35

1$\begingroup$ Do you agree that differential operators on Euclidean space are unbounded? If so, restrict $D$ to a Euclidean neighborhood in $M$. $\endgroup$ – Paul Siegel Sep 8 '14 at 13:53

1$\begingroup$ This is proven in full detail in section 1.6 of the book by Gilkey that you mentioned. $\endgroup$ – Johannes Ebert Sep 8 '14 at 14:59

$\begingroup$ This question appears to be offtopic because it is answered by reading the text mentioned in the question $\endgroup$ – Yemon Choi Sep 8 '14 at 15:08
The fact that (nontrivial) elliptic operators are not bounded on $L^2$ is a special case of the fact that differential operators generally are not continuous (i.e., not bounded) on $L^2$. This much has little to do with the compactness of the manifold.
Note that such operators do not extend "to the completion of $C^\infty(E)$ to $L^2(E)$", as in the question. Yes, such operators have extensions to distributionvalued operators, but that's not immediately to the point.
The fact that symmetric elliptic differential operators have selfadjoint extensions is also a fairly general fact, not depending on compactness, since Friedrichs' selfadjoint extensions always exist for semibounded operators. (The von Neumann or Krein general classification of selfadjoint extensions of symmetric unbounded operators requires that we check some conditions (equality of dimensions of deficiency indices...) before knowing there's any selfadjoint extension.)
The property of eigenvalues mentioned in the question is much more specific, depending for proof on having compact resolvent, which follows for compact manifolds. That argument basically reduces/compares to the case of flat tori, that is, products of circles, where Fourier series give a straightforward proof of the relevant Rellich compactness assertion.