If $\lambda \in \partial \sigma (T)$ then $T - \lambda$ cannot be surjective.

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.

Solution 1:

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.