TITLE: Why is the spectrum usually defined for operators between Banach spaces?
QUESTION [6 upvotes]: The spectrum of a linear operator
$L: \mathcal{D}(L) \rightarrow \mathcal{X} $
is generally defined for $\mathcal{X}$ a Banach space
(as seen for example wikipedia on link above, or spectral decomposition on wikipedia, or this question or this answer).
Why is this? Why don't we define the spectrum more generally for operators between normed spaces? Where do we need the completeness?

REPLY [4 votes]: Spectal Theory evolved with the thought in mind of expanding functions in eigenfunctions of some operator. Eigenfunctions and eigenvalues make no sense if you consider $L : X\rightarrow Y$ because $Lf=\lambda f$ makes no sense if $X\ne Y$. You can't have discrete eigenvalues and discrete eigenfunction expansions, or approximate eigenvvalues and continuous (integral) eigenfunction expansions because the statements no longer make sense if $X\ne Y$.
J. Dieudonne in A History of Functional Analysis details how Spectral Theory came out of looking at the ordinary differential equations arising out of Fourier's separation of variables. The separation parameter gave an eigenvalue equation
$$
                                 Lf=\lambda f.
$$
For finite intervals one had discrete values $\lambda_{n}$ for which non-zero solutions $f_{n}$ existed, and it was found that these eigenfunctions were mutually 'orthogonal' in an integral sense whenever the eigenvalues were different. This led to a general expansion
$$
                            f = \sum_{n} \frac{(f,f_{n})}{(f_{n},f_{n})}f_{n}
$$
where the 'inner-product' pairing was
$$
                      (f,g) = \int_{a}^{b}f(t)g(t)\,w(t)\,dt
$$
for some weight $w$. Using such expansions greatly simplified the process of solving the original partial differential equations; using such functions amounted to a diagonalization of the operator. (By the way, systematic matrix diagonalization came out of these ODE theories, not the other way around.)