Reasons for $\lambda \mathbb{I} - T$ to be non-invertible.

Question

Given a Hilbert space $\mathcal{H}$ an operator $T\in \mathcal{L}(\mathcal{H})$, it's spectrum is $$\sigma(T):= \{ \lambda \in \mathbb{C} : \lambda \mathbb{I} - T \mbox{ is not invertible } \}.$$ Now to actually compute the elements in the spectrum we find out that there are only 3 reasons for which the operator $\lambda \mathbb{I} - T$ is not invertible:
(i) it is not injective;
(ii) it is not surjective;
(iii) it is injective and surjective, but the "inverse" (which is linear) isn't continuous.
But at this point we construct the whole theory on the fact that the case (iii) is not possible; I found a theorem called "Open Mapping Theorem" which would explain it since it would imply that $T$ is an open map, but I can't find a proof.
Could somebody direct me to one or anyways be able to prove it? Also, if what I asked is a problem, I would still be very happy to read an intuitive explanation of how one could prove it.

Answer 1

What you are looking for is the bounded inverse theorem, which says that a bounded linear bijection between Banach space has bounded inverse. You can find a proof in virtually any book that contains some functional analysis. Here are some examples:

• [1] Rudin, "Real and complex analysis", theorem 5.10 (p100).
• 2 Axler, "Measure, integration and real analysis", theorem 6.83 (p188).
• [3] Conway; "A course in functional analysis", theorem 12.5 (p91)

The book 2 is very accessible and can be accessed online for free. See here.