Continuous Functional Calculus for Self Adjoint and Normal Operators — Examples

The formula $Tx=({\lambda }_n{x}_n)$ defines a linear bounded operator with $\mathrm{\parallel }Tx{\mathrm{\parallel }}_2^2={\sum }_n\mathrm{\mid }{\lambda }_n{x}_n{\mathrm{\mid }}^2\le M^2\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so $\mathrm{\parallel }T\mathrm{\parallel }\le M$ and $T{e}_n={\lambda }_n{e}_n$; moreover $\mathrm{\parallel }T\mathrm{\parallel }\ge {\sup }_n\mathrm{\mid }{\lambda }_n\mathrm{\mid }=M$ because $\mathrm{\parallel }T{e}_n\mathrm{\parallel }=\mathrm{\mid }{\lambda }_n\mathrm{\mid }$, so $\mathrm{\parallel }T\mathrm{\parallel }=M$.

$T$ is normal: $T^{\ast }y=(\overline{{\lambda }_n}{y}_n)$, since $⟨Tx,y⟩={\sum }_n{\lambda }_n{x}_n\overline{y_n}={\sum }_n{x}_n\overline{{\lambda }_n{y}_n}$, and therefore $T^{\ast }T=T{T}^{\ast }$ is the diagonal operator with entries $\mathrm{\mid }{\lambda }_n{\mathrm{\mid }}^2$.

$\sigma (T)=\overline{\{{\lambda }_n\}}$: if $z\mathrm{\notin }\overline{\{{\lambda }_n\}}$ then $\delta :={\inf }_n\mathrm{\mid }z-{\lambda }_n\mathrm{\mid }>0$, the diagonal operator $S$ with entries $1\mathrm{/}(z-{\lambda }_n)$ is bounded with $\mathrm{\parallel }S\mathrm{\parallel }\le 1\mathrm{/}\delta$, and $S(zI-T)=(zI-T)S=I$, so $z\in \rho (T)$; if $z\in \overline{\{{\lambda }_n\}}$ choose $n_k$ with ${\lambda }_{n_k}\to z$, so $\mathrm{\parallel }(T-zI)e_{n_k}\mathrm{\parallel }=\mathrm{\mid }{\lambda }_{n_k}-z\mathrm{\mid }\to 0$, whence $T-zI$ is not bounded below and lies in $\sigma (T)$.

For $f\in C(\sigma (T))$ and each $n$, the basis vector $e_n$ is an eigenvector of the normal operator $T$ at ${\lambda }_n\in \sigma (T)$, so $f(T)e_n=f({\lambda }_n)e_n$.

Hence $\sigma (T)=\overline{\{{\lambda }_n\}}$ and the calculus acts diagonally on the standard basis, as asserted.

$M_t$ is well defined on a.e. classes, linear, bounded with $\mathrm{\parallel }{M}_{t}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$, and self-adjoint with $M_t^2=M_{t^2}$; self-adjointness follows from $⟨M_{t}u,v⟩=\int tu\bar{v}=⟨u,M_{t}v⟩$. For $\lambda \in \mathbb{C}$ one has $M_t-\lambda I=M_{t-\lambda }$.

If $z\mathrm{\notin }[0,1]$ then $g(t)=1\mathrm{/}(z-t)$ is continuous on $[0,1]$ with $\mathrm{\mid }g\mathrm{\mid }\le 1\mathrm{/}\mathrm{dist}(z,[0,1])$, so $M_g$ is bounded and $M_g(zI-M_t)=(zI-M_t)M_g=I$; hence $z\in \rho (M_t)$.

If $z\in [0,1]$ and $0<\delta \le 1$, let $f_{\delta }(t):=(1-\mathrm{\mid }t-z\mathrm{\mid }\mathrm{/}\delta {)}_{+}$ on $[0,1]$, a continuous function vanishing wherever $\mathrm{\mid }t-z\mathrm{\mid }\ge \delta$, and put ${\phi }_{\delta }:=c_{\delta }{f}_{\delta }$ with $c_{\delta }$ chosen so that $\mathrm{\parallel }{\phi }_{\delta }{\mathrm{\parallel }}_2=1$. For $z=0$ or $z=1$ take $0<\delta \le 1$; for $0<z<1$ take $0<\delta \le \min (z,1-z)$. Respectively, the polynomial integrals give $\int f_{\delta }^2=\delta \mathrm{/}3$ or $2\delta \mathrm{/}3$ and $\int (t-z{)}^2{f}_{\delta }^2={\delta }^3\mathrm{/}30$ or ${\delta }^3\mathrm{/}15$ respectively, so the ratio of the second integral to the first is ${\delta }^2\mathrm{/}10$ in both cases. Thus $\mathrm{\parallel }(M_t-zI){\phi }_{\delta }{\mathrm{\parallel }}_2^2={\delta }^2\mathrm{/}10$. A bounded inverse $B$ would imply $1\le \mathrm{\parallel }B\mathrm{\parallel }\delta \mathrm{/}\sqrt{10}$ for all such $\delta >0$, which is impossible; hence $M_t-zI$ is not bounded below and $z\in \sigma (M_t)$.

Steps 2.1 and 2.2 give $\sigma (M_t)=[0,1]$, and $M_t$ is self-adjoint, so the continuous functional calculus for $M_t$ is defined on $C([0,1])$.

For a continuous $f$ the multiplier $M_f$ is well defined on a.e. classes, linear, and bounded with $\mathrm{\parallel }{M}_f\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, by $\int \mathrm{\mid }fu{\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}^2\int \mathrm{\mid }u{\mathrm{\mid }}^2$. Multiplication of multipliers and the unital algebra property of the supplied calculus give $p(M_t)=M_p$ for every complex polynomial $p$. For each $\epsilon >0$, [A4] supplies a polynomial $p$ with $\mathrm{\parallel }f-p{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. The calculus isometry and the multiplier bound give $\mathrm{\parallel }f(M_t)-M_f\mathrm{\parallel }\le \mathrm{\parallel }f(M_t)-p(M_t)\mathrm{\parallel }+\mathrm{\parallel }{M}_p-M_f\mathrm{\parallel }\le 2\mathrm{\parallel }f-p{\mathrm{\parallel }}_{\mathrm{\infty }}<2\epsilon$. Since this holds for every $\epsilon >0$, $f(M_t)=M_f$. In particular the isometry and range conclusions already established for the calculus give $\mathrm{\parallel }{M}_f\mathrm{\parallel }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and range $C^{\ast }(I,M_t)$.

Therefore $\sigma (M_t)=[0,1]$ and the calculus of $M_t$ is multiplication by $f$.

$T^{\ast }=\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$ and $T^{\ast }T=\mathrm{diag}(0,4)$: indeed $⟨T{e}_1,e_1⟩=⟨T{e}_1,e_2⟩=0$, $⟨T{e}_2,e_1⟩=2$, $⟨T{e}_2,e_2⟩=0$, so the adjoint has the displayed matrix and the product is diagonal with entries $\mathrm{\parallel }T{e}_1{\mathrm{\parallel }}^2=0$ and $\mathrm{\parallel }T{e}_2{\mathrm{\parallel }}^2=4$.

$\mathrm{diag}(0,4)$ is self-adjoint and positive, as is $\mathrm{diag}(0,2)$: for $x=x_1{e}_1+x_2{e}_2$ one has $⟨\mathrm{diag}(0,4)x,x⟩=4\mathrm{\mid }{x}_2{\mathrm{\mid }}^2\ge 0$ and $⟨\mathrm{diag}(0,2)x,x⟩=2\mathrm{\mid }{x}_2{\mathrm{\mid }}^2\ge 0$.

$\mathrm{diag}(0,2{)}^2=\mathrm{diag}(0,4)=T^{\ast }T$, so by uniqueness of the positive square root $\mathrm{\mid }T\mathrm{\mid }=(T^{\ast }T{)}^{1\mathrm{/}2}=\mathrm{diag}(0,2)$.

Likewise $\mathrm{diag}(1,4)$ is positive with positive square root $\mathrm{diag}(1,2)$, since $\mathrm{diag}(1,2{)}^2=\mathrm{diag}(1,4)$ and $⟨\mathrm{diag}(1,2)x,x⟩=\mathrm{\mid }{x}_1{\mathrm{\mid }}^2+2\mathrm{\mid }{x}_2{\mathrm{\mid }}^2\ge 0$, uniqueness again identifying the square root.

Hence $\mathrm{\mid }T\mathrm{\mid }=\mathrm{diag}(0,2)$ as claimed, and the positive square root of $\mathrm{diag}(1,4)$ is $\mathrm{diag}(1,2)$.

$S$ is a well-defined bounded linear isometry: for $x={\sum }_n{x}_n{e}_n$ the series $Sx={\sum }_n{x}_n{e}_{n+1}$ converges with $\mathrm{\parallel }Sx{\mathrm{\parallel }}_2^2={\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so $S^{\ast }S=I$ and $\mathrm{\parallel }S\mathrm{\parallel }=1$.

$S{S}^{\ast }=I-P$: for the basis vectors $S^{\ast }{e}_0=0$ and $S^{\ast }{e}_{n+1}=e_n$, so $⟨S{S}^{\ast }{e}_m,e_k⟩=⟨S^{\ast }{e}_m,S^{\ast }{e}_k⟩$ equals $1$ for $m=k\ge 1$ and $0$ otherwise; hence $S{S}^{\ast }$ is the identity on the closed span of $\{e_n:n\ge 1\}=(\mathbb{C}{e}_0{)}^{\perp }$ and vanishes on $\mathbb{C}{e}_0$.

$\mathrm{\mid }S\mathrm{\mid }=I$: since $S^{\ast }S=I$, the identity is positive with square $I=S^{\ast }S$, so by uniqueness of the positive square root $\mathrm{\mid }S\mathrm{\mid }=I$.

Consequently $\ker S=\{0\}$ and $\mathrm{ran}S=(\mathbb{C}{e}_0{)}^{\perp }$, and $S$ is an isometry that is not a coisometry: $S{S}^{\ast }\mathrm{\ne }I$ because $S{S}^{\ast }{e}_0=0$.

The polar partial isometry of $S$ is $U=S$: indeed $S=S\cdot I=S\mathrm{\mid }S\mathrm{\mid }$, $\ker S=\{0\}=\ker U$, and $S$ is a partial isometry, being isometric on $H=(\ker S{)}^{\perp }$ and vanishing on $\ker S=\{0\}$; uniqueness in the polar decomposition identifies it.

The initial space is $(\ker S{)}^{\perp }=H$ and the final space is $\overline{\mathrm{ran}S}=(\mathbb{C}{e}_0{)}^{\perp }$, as asserted.

$J^2=0$ and $J\mathrm{\ne }0$, and $zI-J$ is invertible for every $z\mathrm{\ne }0$ with inverse $z^{-1}(I+z^{-1}J)$, because $(zI-J)z^{-1}(I+z^{-1}J)=z^{-1}(zI+J-J-z^{-1}{J}^2)=z^{-1}(zI-z^{-1}{J}^2)=I$ using $J^2=0$.

$J$ is not normal: $J^{\ast }=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, so $J{J}^{\ast }=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ while $J^{\ast }J=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, and these are different operators.

$0\in \sigma (J)$ because $J$ is not injective: $J{e}_1=0$ while $e_1\mathrm{\ne }0$, so $zI-J$ is not bijective at $z=0$; hence $\sigma (J)=\{0\}$.

The operator $J$ is therefore a nonzero bounded operator whose spectrum is the singleton $\{0\}$ — the property the title calls quasinilpotence — and it is not normal; by the zero-spectrum corollary this is only possible because normality fails.

The statement that nonzero operators have nonzero spectrum, equivalently that zero spectrum forces vanishing, is refuted by the witness $J$; dropping the normality hypothesis from the zero-spectrum corollary is therefore not legitimate.

$P$ is an orthogonal projection commuting with $T$: pointwise multiplication by an indicator is idempotent and self-adjoint, and multiplication operators commute, so $PT=TP$.

If $P=f(T)$ for a continuous $f\in C([0,1])$, then $M_f=P=M_{{\mathbf{1}}_{(0,1\mathrm{/}2)}}$ by the multiplication form of the calculus, so the two bounded functions agree as elements of $L^2(0,1)$, that is $f={\mathbf{1}}_{(0,1\mathrm{/}2)}$ almost everywhere.

A continuous function agreeing almost everywhere with an indicator of a proper subinterval is impossible: $f=1$ a.e. on $(0,1\mathrm{/}2)$ and $f=0$ a.e. on $(1\mathrm{/}2,1)$, so continuity at $1\mathrm{/}2$ would give $f(1\mathrm{/}2)={\lim }_{t\to 1\mathrm{/}{2}^{-}}f(t)=1$ and simultaneously $f(1\mathrm{/}2)={\lim }_{t\to 1\mathrm{/}{2}^{+}}f(t)=0$; the two limits are computed along intervals where the continuous function is a.e. constant, hence constant there.

Hence no continuous $f$ satisfies $P=f(T)$, so the commuting projection $P$ demonstrates that the continuous calculus does not contain every spectral projection of $T$.

$J$ has real spectrum $\{0\}\subseteq [0,+\mathrm{\infty })$: $J^2=0$ gives the inverse $z^{-1}(I+z^{-1}J)$ of $zI-J$ for every $z\mathrm{\ne }0$, while $z=0$ is a spectral value because $J{e}_1=0$.

$J$ is not positive: for $x:=\frac{1}{\sqrt{2}}(e_1+i{e}_2)$ one has $Jx=\frac{i}{\sqrt{2}}{e}_1$ and hence $⟨Jx,x⟩=\frac{i}{\sqrt{2}}\cdot \overline{\frac{1}{\sqrt{2}}}=\frac{i}{2}$, which is not a real number, while positivity of a bounded operator requires the value of the quadratic form at every vector to be a real number in $[0,+\mathrm{\infty })$.

$J$ is not self-adjoint: the defining pairing on the standard orthonormal basis gives $J^{\ast }{e}_1=e_2$ and $J^{\ast }{e}_2=0$, so $J^{\ast }=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\mathrm{\ne }J$.

The witness $J$ therefore has nonnegative real spectrum but is neither positive nor self-adjoint, so nonnegativity of the spectrum alone does not give the quadratic-form inequalities of the order calculus.

The statement is refuted: $J$ satisfies its spectral antecedent but fails its quadratic-form conclusion, so spectral nonnegativity alone cannot extend the self-adjoint order definition to all bounded operators.