Consider the Hilbert L^2([0,1]) associated with a physical particle that has position somewhere between 0 and 1, the corresponding multiplication operator X which takes a wavefunction f and maps it to Xf where (Xf)(x) = x f(x). Then X is a bounded self-adjoint operator. It doesn't have any eigenvalues or eigenvectors but it's spectrum is exactly the set of numbers [0,1] as you'd expect (prefect measurements of position return real numbers in [0,1]).
The spectral theorem, rather than decomposing X in terms of a sum of eigenvectors & eigenvalues instead decomposes it as an integral over the spectrum with respect to the (spectral) projection-valued measure.
Now it is fair to question whether this "observable" is really observable, but it certainly works out mathematically consistently in the normal way we do things in quantum mechanics.
The spectral theorem, rather than decomposing X in terms of a sum of eigenvectors & eigenvalues instead decomposes it as an integral over the spectrum with respect to the (spectral) projection-valued measure.
Now it is fair to question whether this "observable" is really observable, but it certainly works out mathematically consistently in the normal way we do things in quantum mechanics.