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If I have a normal bounded operator $T$ on a (say separable) Hilbert space $H$ then the following resolvent growth condition holds:

$||R(z,T)||=\frac{1}{dist(z,\sigma(T))}=r(R(z,T))$

where $R(z,T)=(T-zI)^{-1}$ is the resolvent and $\sigma(T)$ denotes the spectrum of $T$ and $r$ the spectral radius. In the non-normal case things are not so pleasant and we may have

$||R(z,T)||>\frac{1}{dist(z,\sigma(T))}$

(note that we always have $\geq$ in the above). My question is whether there are known cases (with $H$ infinite dimensional and $T$ not compact) where growth of $R(z,T)$ is known in terms of $dist(z,\sigma(T))$ which isn't just the linear case above. For instance, replacing the linear growth by quadratic growth?

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  • $\begingroup$ Why does the equality hold for normal operators? I can see it for self adjoint, but not sure why it works for a general normal operator $\endgroup$
    – Niebla
    Dec 18 '17 at 7:29
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Start with a matrix on $\mathbb{C}^2$ (you can extend to a non-compact operator in $\ell^2$ later) defined as $$ A = \left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\0 & 0 & 0\end{array}\right]. $$ Then $A^3=0$ and $A^2 \ne 0$, as $A$ maps the third basis element to the second, the second to the first, and the first to the zero vector. You can directly verify that the resolvent of $A$ is defined for all $\lambda\ne 0$ by $$ (A-\lambda I)^{-1}=-\frac{1}{\lambda^3}(A^2+\lambda A + \lambda^2 I) $$ Near $\lambda=0$, the resolvent has a $1/\lambda^3$ behavior. For example, if you apply the above to the third basis vector, you $-1/\lambda^3$ times the first basis vector plus $-1/\lambda^2$ times the second basis vector and $-1/\lambda$ times the third basis vector. You can certainly operator norm bounded $(A-\lambda I)^{-1}$ by a constant times $1/|\lambda|^3$, and that order of bound is achieved.

Finally, you can put this into $\ell^2$ in such a way that the result is not compact. To do this, define $A$ on the first three basis vectors, and let $A$ be the identity on the subspace spanned by the remaining basis vectors. Then $A$ is not compact. The resolvent of the extended operator is defined everywhere except for $\{0,1\}$. If you use a higher-order nilpotent matrix on a higher-dimensional subspace, you end up with a higher-order pole of the resolvent. That way you can get any reciprocal polynomial behavior you want for the resolvent.

Why this is general: This is basically the general case for an isolated point of the spectrum. For example, if $\mu$ is an isolated point of the spectrum, then you can choose any positively oriented circular contour in the resolvent set that encloses only $\mu$ from the spectrum in its interior, and then show that the following is a projection $$ P_{\mu} = \frac{1}{2\pi i}\oint_{C} \frac{1}{\lambda I-A}d\lambda. $$ Then, if you can choose $n$ large enough that $(\lambda-\mu)^{n}(\lambda I-A)^{-1}$ is non-singular, you obtain a nilpotent condition: $$ (A-\mu I)^n P_{\mu} = \frac{1}{2\pi i}\oint_{C}\frac{(\lambda-\mu)^{n}}{\lambda I-A}d\lambda = 0 $$ In other words, $A-\lambda I$ is nilpotent of order $n$ on $P_{\mu}X$ iff $(A-\lambda I)^{-1}$ has an isolated singularity of order $n$ at $\mu$. I gave you an example of this, but it's basically the general case for an isolated point of the spectrum. (You can use this in finite-dimensional spaces to derive the Jordan blocks associated with $\mu$ because the singularities of the resolvent are all isolated singularities of finite order in finite-dimensional cases.)

It should be noted that in infinite-dimensional spaces, the resolvent can have an isolated essential singularity (i.e., of infinite order.)

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  • $\begingroup$ Nice! But this seems like cheating a bit since we can split the operator in a trivial way. $\endgroup$
    – Mathmo
    Sep 7 '16 at 18:54
  • $\begingroup$ @Mathmo : I added more so that you can see why this is more general than you might think. $\endgroup$ Sep 7 '16 at 19:32
  • $\begingroup$ Ah, I see yes. Yes it's making more sense. Are there conditions that ensure we don't have an isolated essential singularity? For instance, compact perturbations of normal operators? $\endgroup$
    – Mathmo
    Sep 9 '16 at 7:10
  • $\begingroup$ @Mathmo : If a normal operator has a resolvent with an isolated singularity at $\lambda$, then the order of the pole is 1 because $(\lambda I-A)^{-2}P_{\mu}=0$ implies $(\lambda I- A)^{-1}P_{\mu}=0$. I'll let you consider the Volterra operator $Vf = \int_{0}^{x}f(t)dt$ on $C[0,1]$ or on $L^2[0,1]$. $V$ is compact, which makes it a compact perturbation of $0$. Try it, and see what you get for the resolvent. $\endgroup$ Sep 9 '16 at 12:31

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