4
$\begingroup$

If we have a Banach space $X$ and we consider the space $L(X,X)$ of linear operators. Now we have the operator norm here and this induces a metric, which in turn induces the topology. Since this is a metric space, we have the notion of open sets. So how would one show that a certain subspace is open? I'm having difficulty seeing what the open balls look like.

So the question from Folland says that: if $T$ is invertible, and $||S-T||\leq||T^{-1}||^{-1}$, then S is invertible. He then concludes that the set of Invertible Operators is open. I dont see how this follows, could someone explain please?

$\endgroup$
5
  • $\begingroup$ It's hard to see what an open ball looks like in infinite dimension. For an example, the set of invertible operators in $L(X,X)$ is open. $\endgroup$ Commented Nov 23, 2015 at 4:33
  • $\begingroup$ Thats exactly what i want to show but im not sure that im looking at it from the right way, the topology generated by metric balls way. $\endgroup$
    – Atherton
    Commented Nov 23, 2015 at 4:37
  • 1
    $\begingroup$ Even though we can't really visualize it, the definition of open set stays the same: you need to show that there exists some positive number $r$ such that if $T$ is invertible and $\|S-T\|<r$ then $S$ is invertible as well. This is just saying that around any element of the open set there exists an open ball which is all contained in the open set. $\endgroup$ Commented Nov 23, 2015 at 4:39
  • $\begingroup$ ok ill get cracking now $\endgroup$
    – Atherton
    Commented Nov 23, 2015 at 4:51
  • $\begingroup$ Just to make sure that what im doing is right. Lets take $T,S\in L(X,X)$, with $T$ invertible. Then(im not sure about this inequality as it relies on $||-T||=||T||$) look at $||S-T||\leq ||S||+||T||$ but the things on the right were $||S||=inf\{c: ||Tx|| \leq c||x||\}$ for all x. Hence, $||S-T||\leq ||S||+||T|| \leq \alpha +\beta$ where $\alpha$ and $\beta$ are the corresponding infimums. Hence, we show that there exists an $r=\alpha+\beta$ such that $||S-T||\leq r$. $\endgroup$
    – Atherton
    Commented Nov 23, 2015 at 5:17

3 Answers 3

6
$\begingroup$

This is a late answer, but I feel the current answers are unsatisfying and I had the same question for a bit when I first encountered this myself so it seems worth explaining with a bit more care.

Recall the definition of an open set as set where every point is an interior point. Denote the set of invertible linear operators by $\Omega$. So, what we wish to show is that if $T \in \Omega$, then there exists a ball $B_\delta(T)$ such that $B_\delta(T) \subset \Omega$. That is, there exists $\delta > 0$ such that $\Vert S- T \Vert < \delta$ implies $S \in \Omega$.

The claim in Folland's book asserts that for any $S \in L(X,X)$ which satisfies $\Vert S-T \Vert \leq \Vert T^{-1}\Vert^{-1}$, we have that $S \in \Omega$. So, taking $\delta < \Vert T^{-1} \Vert^{-1}$ gives us exactly the condition that $B_\delta(T) \subset \Omega$.

It turns out that this is extremely easy to prove straight from the definitions, just it can be a bit hard to see how to unpack them when dealing with unfamiliar objects (such as linear operators in a Banach space instead of points in $\mathbb{R}^n$). The real lesson here is just the trivial statement that definitions apply everywhere they apply, not just the intuitive/familiar cases.

$\endgroup$
5
$\begingroup$

Writing $S = T + W = T (I + T^{-1} W)$. It suffices to show $I + T^{-1} W$ is invertible, where $\|T^{-1} W\| \le \|T^{-1}\| \|W\| < 1$. Write $(I + T^{-1} W)^{-1}$ as a series...

$\endgroup$
0
$\begingroup$

In general this depends on what norm you are using, and what space you are on. For example, let's use the norm

\begin{align} \|A\| = \sup_{\|f\|_X=1}\|Af\|_X \end{align}

Where the norms on the right are taken in $X$ and if $X = \mathbb{R}^n$, then $L(X)$ is isomorphic to $\mathbb{R}^{n^2}$(as we are just looking at matricies), with the "taxicab norm". With this norm the balls will look a lot like boxes. But we could easily choose another matrix norm and get another shape for the open sets. And this is only for the finite dimensional case, when we get to the infinite dimensional case, we have a lot more possibilities for operator norms. Also in the infinite dimensional case, there are topologies that are not induced by norms, for example the weak toplogy, which will have open sets that look different as well.

So if one wanted to show a set is open, one would first need to decide on a norm (or other way to define an open set) and use the tools from topology to find these. For example you could show that the complement is closed , or that for every point, there is an open ball around that point that stays in the set. But what all of these notions mean in this specific case is determined by the space and way that you define a topology.

$\endgroup$
1
  • 1
    $\begingroup$ Unless otherwise specified, the norm for $L(X,X)$, where $X$ is a Banach space, is the operator norm. $\endgroup$ Commented Nov 23, 2015 at 6:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .