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Let $\mathcal H$ be a Hilbert space and $T \in \mathcal L(\mathcal H).$ If $\lambda \in \partial \sigma (T)$ then $T - \lambda$ is not surjective.

This question appeared in an entrance examination in India for admission into PhD programme which I am unable to solve. Any hint would be a boon for me at this stage.

Thanks for your time.

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  • $\begingroup$ What's the notation $\partial\sigma(T)$ mean? $\endgroup$
    – Chee Han
    Dec 23, 2021 at 16:44
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    $\begingroup$ @Chee Han$:$ The boundary of the spectrum of $T.$ $\endgroup$
    – RKC
    Dec 23, 2021 at 16:46
  • $\begingroup$ Then I am confused. The spectrum of a bounded linear operator is closed, which means $\lambda$ is in the spectrum, which, by definition, means that the operator $T - \lambda I$ is not bijective. $\endgroup$
    – Chee Han
    Dec 23, 2021 at 16:53
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    $\begingroup$ @Chee Han$:$ Yes that's correct. But what's the source of confusion? If we take any point from the boundary of the spectrum then the corresponding operator can't be surjective and hence not bijective as well. $\endgroup$
    – RKC
    Dec 23, 2021 at 16:58
  • $\begingroup$ Wow ok, I see my flaws now. Wow, this is an interesting question. $\endgroup$
    – Chee Han
    Dec 23, 2021 at 17:21

2 Answers 2

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Here's a sketch, I think.

Lemma 1.

If $(A_n)_n$ is a sequence in $\mathcal{L}(\mathcal{H})$ of invertible operators such that $A_n \to A$ where $A\in \mathcal{L}(\mathcal{H})$ is not invertible, then $A$ is not bounded from below.

In our case we can find a sequence $(\lambda_n)_n$ in $\sigma(T)^c$ such that $\lambda_n \to \lambda$. Then $T-\lambda_n I \to T-\lambda I$ but since $T - \lambda I$ is not invertible, it also isn't bounded from below.

Lemma 2.

For $A \in \mathcal{L}(\mathcal{H})$ we have that $A$ is bounded from below if and only if $A^*$ is surjective.

Hence $(T-\lambda I)^* = T^* - \overline{\lambda} I$ is not surjective. Not quite what we wanted, but if $\lambda \in \partial \sigma(T)$, then $\overline{\lambda} \in \partial\sigma(T^*)$ so we can apply the above proof to $T^* - \overline{\lambda} I$ to conclude that its adjoint, which is $T- \lambda I$, is not surjective.


To prove Lemma 1 we first need Lemma 0:

If $(A_n)_n$ is a sequence in $\mathcal{L}(\mathcal{H})$ of invertible operators such that $A_n \to A$ where $A\in \mathcal{L}(\mathcal{H})$ is not invertible, then the sequence $(\|A_n^{-1}\|)_n$ is unbounded.

Indeed, assume that there exists $M>0$ such that $\|A_n^{-1}\| \le M$ for all $n \in \Bbb{N}$. Pick $n \in \Bbb{N}$ large enough so that $\|A_n-A\| < \frac1M$. Then we have $$\|A_n^{-1}A-I\| = \|A_n^{-1}(A-A_n)\| \le \|A_n^{-1}\|\|A-A_n\| < M \cdot \frac1M = 1$$ so it follows that $A_n^{-1}A$ is invertible. Hence $A$ is also invertible, which is a contradiction.

(In fact, by passing to a subsequence we easily see that $\lim_{n\to\infty} \|A_n^{-1}\| = +\infty$.)

Now onto the proof of Lemma 1. Assume that $A$ is bounded from below. Let $m> 0$ be such that $\|Ax\| \ge m\|x\|$ for all $x \in \mathcal{H}$.

Then for all $x \in \mathcal{H}$ and $n \in \Bbb{N}$ we have $$\|A_n x\| \ge \underbrace{\|Ax\|}_{\ge m\|x\|} - \underbrace{\|(A_n-A)x\|}_{\le \|A_n-A\|\|x\|} \ge (m-\|A_n-A\|)\|x\|$$ and therefore $$\|x\| = \|A_nA_n^{-1}x\| \ge (m-\|A_n-A\|)\|A_n^{-1}x\| \implies \|A_n^{-1}x\| \le \frac{1}{m-\|A_n-A\|}\|x\|.$$ Since $x \in \mathcal{H}$ was arbitrary, it follows that $$\|A_n^{-1}\| \le \frac{1}{m-\|A_n-A\|}, \quad \forall n \in \Bbb{N}$$ and hence $$\limsup_{n\to\infty} \|A_n^{-1}\| \le \limsup_{n\to\infty} \frac{1}{m-\|A_n-A\|} = \frac1m < +\infty$$ which contradicts Lemma 0. Therefore, $A$ is not bounded from below.

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  • $\begingroup$ How do I show lemma 1? What will happen if $A$ is bounded below? $\endgroup$
    – RKC
    Dec 23, 2021 at 18:57
  • $\begingroup$ Lemma 2 is rather easy. $A$ is bounded below $\iff$ $A$ is injective and $\text {ran}\ A$ is closed $\iff$ $A$ is injective and $\text {ran}\ A^*$ is closed $\iff$ $\text {ran}\ A^*$ is closed and dense $\iff$ $A^*$ is surjective. Done! $\endgroup$
    – RKC
    Dec 23, 2021 at 19:27
  • $\begingroup$ @RabinKumarChakraborty I've added the proof of Lemma 1, we first need another lemma. $\endgroup$ Dec 23, 2021 at 23:13
  • $\begingroup$ Could you share how you came up with the idea of your proof? My first attempt was to argue by contradiction and it led me nowhere, mainly because I don't have an intuition of what would went wrong if $T - \lambda I$ is not injective. $\endgroup$
    – Chee Han
    Dec 24, 2021 at 5:47
  • $\begingroup$ @CheeHan This was my first attempt too. Namely, for $A \in \mathcal{L}(\mathcal{H})$ we have that $A$ is invertible if and only if $A$ is bounded from below and surjective. So, if $T - \lambda I$ were surjective, since it isn't invertible it would follow that it isn't bounded from below. But this isn't a contradiction, in fact it is true by Lemma 1. Then I remembered Lemma 2 and concluded that $(T-\lambda I)^*$ is not surjective, which was similar to what we wanted to show. $\endgroup$ Dec 24, 2021 at 22:33
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An alternative proof to the accepted answer. But first a lemma relating a bounded operator with its adjoint.

Lemma. For Banach spaces $X$ and $Y$, and $A\in\mathcal{B}(X;Y)$, let $A^*: Y^*\to X^*$ be its adjoint, defined as $A^*: f\mapsto f\circ A$. Then

$A^*$ is surjective iff $A$ is bounded below;
$A^*$ is injective iff $A$ is surjective.

Combining the two results, $A^*$ is invertible iff $A$ is invertible.

A proof for this lemma can be found in standard textbooks. Next, the proof for the OP's question.

Proof. Denote $A_\lambda:= T-\lambda$. Then $\text{Ker}(A_\lambda)$ is closed, $\mathcal{H}=\text{Ker}(A_\lambda)\oplus \text{Ker}(A_\lambda)^\perp$, and the restricted map $A_\lambda|_{\text{Ker}(A_\lambda)^\perp}: \text{Ker}(A_\lambda)^\perp\to \mathcal{H}$ is an injective bounded operator.

Since $\lambda\in \partial(\sigma(T))$, and since the spectrum is closed and contains all its boundary points, $A_\lambda$ cannot be invertible. Therefore, if $A_\lambda$ is injective, then $A_\lambda$ cannot be surjective and we are done.

Suppose $A_\lambda$ is not injective. Then $\text{Ker}(A_\lambda)\neq \{0\}$. Assume $A_\lambda$ is surjective. Then $A_\lambda|_{\text{Ker}(A_\lambda)^\perp}$ is invertible, and, by the lemma, so is $\left(A_\lambda|_{\text{Ker}(A_\lambda)^\perp}\right)^*\in \mathcal{B}\left(\mathcal{H}; \text{Ker}(A_\lambda)^\perp\right)$.

Next, consider a sequence of invertible operators $A_n$ converging to $A_\lambda$ in operator norm. Then the restricted maps $A_n|_{\text{Ker}(A_\lambda)}$ and $A_n|_{\text{Ker}(A_\lambda)^\perp}$ are both bounded and bounded below, with $$\mathcal{H}=\text{Ran}(A_n)=\text{Ran}(A_n|_{\text{Ker}(A_\lambda)})\oplus \text{Ran}(A_n|_{\text{Ker}(A_\lambda)^\perp})\supsetneq\text{Ran}(A_n|_{\text{Ker}(A_\lambda)^\perp}).$$ That is, for each $n\in\mathbb{N}$, the restricted map $A_n|_{\text{Ker}(A_\lambda)^\perp}$ is not surjective. Then, by the lemma above, every adjoint $\left(A_n|_{\text{Ker}(A_\lambda)^\perp}\right)^*\in \mathcal{B}\left(\mathcal{H}; \text{Ker}(A_\lambda)^\perp\right)$ is bounded surjective but not injective. By norm convergence of the restricted maps $A_n|_{\text{Ker}(A_\lambda)^\perp}\to A_\lambda|_{\text{Ker}(A_\lambda)^\perp}$, we also have norm convergence of the non-invertible adjoints $\left(A_n|_{\text{Ker}(A_\lambda)^\perp}\right)^*\to \left(A_\lambda|_{\text{Ker}(A_\lambda)^\perp}\right)^*$. But the set of non-invertible operators in $\mathcal{B}\left(\mathcal{H}; \text{Ker}(A_\lambda)^\perp\right)$ is norm closed, and contains all of its limit points. Hence $\left(A_\lambda|_{\text{Ker}(A_\lambda)^\perp}\right)^*$, the norm limit of a non-invertible sequence, is not invertible. Contradiction. $\square$

The last part of this answer is credited to the comment by @mechanodroid under this question.

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