Does $\sigma(T) = \{1\}$ and $\|T\| = 1$ imply that $T$ is the identity?

Solution 1:

The answer is no, it does not have to be the identity even in Hilbert space. One counterexample can be constructed as follows.

Let $H^2$ be the Hardy space in the unit disc and $S$ be the unilateral shift, i.e. $Sf=zf$, $f\in H^2$. Then the classical functional model for a completely nonunitary contraction (aka class $C_0$) is $T=P_KS|K$ where $K=H^2\ominus\Theta H^2$, $\Theta$ is an inner function and $P_K$ is the orthogonal projection on $K$. It is the known fact that $\sigma(T)$ is the spectrum of $\Theta$ (zeros and limiting zeros). If we take the singular inner function $$ \Theta_s(z)=\exp\left(\frac{z+1}{z-1}\right) $$ then the only point of spectrum becomes $z=1$. The norm inequality for the contraction $\|T\|\le 1$ becomes equality as the spectral radius is $1$. It is also known in the literature as a unicellular operator or a Jordan block.

More about the functional model can be found in

• [1] Sz.-Nagy, Foias, "Harmonic Analysis of Operators in Hilbert Space" (Ch.III),
• [2] Nikol'skii, "Treatise on the Shift Operator",
• [3] Bercovici, "Operator Theory and Arithmetic in $H^\infty$".

In particular, in [3,Ch.IV.3] one can find another example. Let $$ Vf(x)=\int_0^x f(t)\,dt,\quad f\in L^2[0,1], $$ (it is a Volterra operator, i.e. $\sigma(V)=\{0\}$) and $T=(I-V)(I+V)^{-1}$. Then $T$ is unitary equivalent to $P_K\Theta_s|K$ above. Alternatively, this $T$ is the cogenerator of a strongly continuous semigroup of contractions ${\cal T}(t)$ that vanishes after $t=1$ (see [3]).