example of bounded operator with no eigenvalues

In this entry we show that there are operators with no eigenvalues. Moreover, we exhibit an operator $T$ in a Hilbert space which is bounded, self-adjoint, has a non-empty spectrum but no eigenvalues.

Consider the Hilbert space $L^2([0,1])$ (http://planetmath.org/L2SpacesAreHilbertSpaces) and let $f:[0,1]⟶ℂ$ be the function $f(t)=t$.

Let $T:L^2([0,1])⟶L^2([0,1])$ be the operator of multiplication (http://planetmath.org/MultiplicationOperatorOnMathbbL22) by $f$

$$T(\phi )=f\phi ,\phi \in L^2([0,1])$$

Thus, $T$ is a bounded operator, since it is a multiplication operator (see this entry (http://planetmath.org/OperatorNormOfMultiplicationOperatorOnL2)). Also, it is easily seen that $T$ is self-adjoint.

We now prove that $T$ has no eigenvalues: suppose $\lambda \in ℂ$ is an eigenvalue of $T$ and $\phi$ is an eigenvector. Then,

$$T\phi =\lambda \phi$$

This means that $(f-\lambda )\phi =0$, but this is impossible for $\phi \ne 0$ since $f-\lambda$ has at most one zero. Hence, $T$ has no eigenvalues.

Of course, since the Hilbert space is complex, the spectrum of $T$ is non-empty (see this entry (http://planetmath.org/SpectrumIsANonEmptyCompactSet)). Moreover, the spectrum of $T$ can be easily computed and seen to be the whole interval $[0,1]$, as we explain now:

It is known that an operator of multiplication by a continuous function $g$ is invertible if and only if $g$ is invertible. Thus, for every $\lambda \in ℂ$, $T-\lambda I$ is easily seen to be the operator of multiplication by $(f-\lambda )$. Hence, $T-\lambda I$ is not invertible if and only if $\lambda \in [0,1]$, i.e. $\sigma (T)=[0,1]$.