Elliptic operator are unbounded [closed]

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 off-topic by Chris Gerig, Michael Renardy, Johannes Ebert, Stefan Kohl, Yemon ChoiSep 8 '14 at 15:08

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• 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. – Paul Siegel Sep 8 '14 at 13:06
• Also could you please point out why the operator will be unbounded. Thanks. – ankit Sep 8 '14 at 13:35
• Do you agree that differential operators on Euclidean space are unbounded? If so, restrict $D$ to a Euclidean neighborhood in $M$. – Paul Siegel Sep 8 '14 at 13:53
• This is proven in full detail in section 1.6 of the book by Gilkey that you mentioned. – Johannes Ebert Sep 8 '14 at 14:59
• This question appears to be off-topic because it is answered by reading the text mentioned in the question – Yemon Choi Sep 8 '14 at 15:08

The fact that (non-trivial) 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 distribution-valued operators, but that's not immediately to the point.