Prove $\lambda(\lambda - T)^{-1} \rightarrow \chi_{\{0\}}(T)$ as $\lambda \rightarrow 0$

Question

The following question is from Linear Operators edited by Nelson Dunford and Jacob T. Schwartz, Chapter VII.8, Problem 7.

Given a banach space $X$ and a bounded operator $T \in B(X)$, we assume $0$ is an isolated point in $\sigma(T)$ and also a pole of the function $(\lambda - T)^{-1}$ that is analytic in a neighborhood of $0$ (but not analytic at $0$). Assume there exists $K \in \mathbb{R}$ such that for small enough $\epsilon > 0$ we have $\,\|\lambda\,(\lambda - T)^{-1}\,\| \leq K\,\forall\,\vert\lambda\vert \in (0, \epsilon)$. Then show that $\lambda\,(\lambda - T)^{-1}$ converge to the projection $E[0] = \chi_{\{0\}}(T)$ in operator norm as $\lambda$ approaches $0$.

This is my attempt. Since $\,\|\lambda^2(\lambda - T)^{-1}\,\|\rightarrow 0$ as $\lambda \rightarrow 0$, then if let $C_{\epsilon}$ be the boundary of the open disk $\{\vert\lambda\vert < \epsilon\}$, then we have the norm of $\int_{C_{\epsilon}}\lambda^2(\lambda -T)^{-1}d\lambda = T^2E[0]$ can be arbitraily small as $\epsilon$ approaches $0$ and hence $T^2E[0] = 0$. Then the order of $0$ as a pole of $(\lambda - T)^{-1}$ is at most $2$. Since $\|\,\lambda(\lambda - T)^{-1}\,\|$ is bounded near $0$ the order of $0$ as a pole of $(\lambda - T)^{-1}$ can only be $1$. Then I cannot proceed. If the conclusion is true then $TE[0]$ is indeed $0$ but I do not know how to prove it...

Any hints will be appreciated. Now for a pole $\lambda$ of an operator $T$ (an isolated point in $\sigma(T)$ and a pole of the function $(\lambda - T)^{-1}$ that is analytic near $\lambda$, let the least integer $n$ such that $(\lambda - T)^n\,E[\lambda] = 0$ while $(\lambda - T)^{n-1}\,E[\lambda] \neq 0$ be the order of a pole(here is $\lambda$). If we let $g$ be a $B(X)$-valued function defined on $\mathbb{C}$, analytic near a complex number $\lambda$ but has it as a pole of order $n$, can I write $g(x) = \frac{1}{(x - \lambda)^n}f(x)$ when $x$ is near $\lambda$?

Answer 1

Once you know that you have a single-order pole, then $$ TE[0]=\frac{1}{2\pi i}\int_{C} \lambda(\lambda-T)^{-1}d\lambda = 0. $$ This is because you can shrink the contour around $\lambda=0$, while $\lambda(\lambda-T)^{-1}$ remains bounded, the length of the contour tends to $0$.

Answer 2

Letting $\varepsilon $ be such that $\overline{B_\varepsilon (0)}\cap \sigma (T)=\{0\}$, set $$ U=\{z\in \mathbb C: |z|\neq \varepsilon \}, $$ and for each $\lambda \neq 0$, with $|\lambda |<\varepsilon $, let $f_\lambda $ be the function defined on $U$ by $$ f_\lambda (z) = \left\{\matrix{ \displaystyle {\lambda \over \lambda -z}, & \text{ if } |z|>\varepsilon , \cr 1, & \text{ if } |z|<\varepsilon .}\right. $$ Notice that, as $\lambda \to 0$, $f_\lambda $ converges to the characteristic function of $\{0\}$ on every compact subset of $U$. Letting $$ \Psi:\mathscr H(U)\to B(H), $$ be the algebra homomorphism behind the holomorphic functional calculus, we then have that $$ \lim_{\lambda \to 0} \Psi(f_\lambda ) = E[0]. $$ However, since $f_\lambda ={\lambda \over \lambda -z}$ on the spectrum of $T$, we have that $ \Psi(f_\lambda ) = \lambda(\lambda -T)^{-1}, $ so $$ \lim_{\lambda \to 0} \lambda(\lambda -T)^{-1} = E[0]. $$

Answer 3

This is an answer for the last question in the original post.

Let $E$ be a Banach space, let $z_0\in {\mathbb C}$, and let $U$ be any open neighborhood of $z_0$. Let us also fix a holomorphic function $$ f:U\setminus\{z_0\}\to E. $$

Finally, let us fix any closed path $\gamma $ in $U$, winding around $z_0$ counter-clockwise once, such as $\gamma (t) = z_0+re^{it}$, for a sufficiently small $r$.

Lemma. Suppose that $\varphi \circ f$ has a removable singularity at $z_0$, for every $\varphi $ in the topological dual $E'$ of $E$. Then $f$ has a removable singularity at $z_0$, as well.

Proof. Given $\varphi $ in $E'$, let $g_\varphi $ be the unique complex valued holomorphic function defined on the whole of $U$, such that $$ g_\varphi (z) = \varphi (f(z)), \quad \forall z\in U\setminus\{z_0\}. \tag{1} $$

Observe that by Cauchy's integral formula we have $$ g_\varphi (z_0) = {1\over 2\pi i}\int_\gamma {g_\varphi (z)\over z-z_0}\,dz = $$ $$= {1\over 2\pi i}\int_\gamma {\varphi (f(z))\over z-z_0}\,dz= $$ $$= \varphi \left({1\over 2\pi i}\int_\gamma {f(z)\over z-z_0}\,dz\right). \tag{2} $$ Extending $f$ to $U$ by setting $$ f(z_0)={1\over 2\pi i}\int_\gamma {f(z)\over z-z_0}\,dz, $$ we then have by (2) that $ g_\varphi (z_0) = \varphi (f(z_0)) $ and, with (1), we conclude that $g_\varphi =\varphi \circ f$ on the whole of $U$.

Since every $g_\varphi $ is holomorphic, we have that the extended function $f$ is weakly holomorphic, and hence holomorphic, as desired. QED

Theorem. For each (positive or negative) integer $n$, let $$ u_n= {1\over 2\pi i}\int_\gamma {f(z)\over (z-z_0)^{n+1}}\,dz, $$ and suppose that $u_n=0$, for every $n<n_0$, where $n_0$ is a (possibly negative) fixed integer. Then the function $F$ defined on $U\setminus\{z_0\}$ by $$ F(z) = {f(z)\over (z-z_0)^{n_0}} $$ has a removable singularity at $z_0$.

Proof. Given any $\varphi $ in $E'$, let $g_\varphi =\varphi \circ f$, and observe that the Laurent series representing $g_\varphi $ on $U\setminus\{z_0\}$ is given by $$ g_\varphi (z) = \sum_{n=-\infty }^\infty \varphi (u_n)(z-z_0)^n = \sum_{n=n_0}^\infty \varphi (u_n)(z-z_0)^n. $$ It then follows that the function $h_\varphi $ defined on $U\setminus\{z_0\}$ by $$ h_\varphi (z) ={g_\varphi (z)\over (z-z_0)^{n_0}}, $$ has a removable singularity at $z_0$.

Observing that $h_\varphi $ coincides with $\varphi \circ F$, the conclusion follows immediately from the Lemma. QED