example of bounded operator with no eigenvalues
Consider the Hilbert space (http://planetmath.org/L2SpacesAreHilbertSpaces) and let be the function .
Let be the operator of multiplication (http://planetmath.org/MultiplicationOperatorOnMathbbL22) by
We now prove that has no eigenvalues: suppose is an eigenvalue of and is an eigenvector. Then,
This means that , but this is impossible for since has at most one zero. Hence, has no eigenvalues.
Of course, since the Hilbert space is complex, the spectrum of is non-empty (see this entry (http://planetmath.org/SpectrumIsANonEmptyCompactSet)). Moreover, the spectrum of can be easily computed and seen to be the whole interval , as we explain now:
It is known that an operator of multiplication by a continuous function is invertible if and only if is invertible. Thus, for every , is easily seen to be the operator of multiplication by . Hence, is not invertible if and only if , i.e. .
|Title||example of bounded operator with no eigenvalues|
|Date of creation||2013-03-22 17:57:53|
|Last modified on||2013-03-22 17:57:53|
|Last modified by||asteroid (17536)|