Continuity of the inverse of a linear operator

[AxiomOfChoice]

If [itex]g(a) \neq 0[/itex] and both [itex]f[/itex] and [itex]g[/itex] are continuous at [itex]a[/itex], then we know the quotient function [itex]f/g[/itex] is continuous at [itex]a[/itex].

Now, suppose we have a linear operator [itex]A(t)[/itex] on a Hilbert space such that the function [itex]\phi(t) = \| A(t) \|[/itex], [itex]\phi: \mathbb R \to [0,\infty)[/itex], is continuous at [itex]a[/itex]. Do we then know that the function [itex]\varphi(t) = \|A(t)^{-1}\|[/itex], [itex]\varphi: \mathbb R \to [0,\infty)[/itex] is continuous at [itex]a[/itex], provided the inverse exists there? Any ideas on how to tackle this question?

I guess I should add that [itex]A(t)[/itex] is a family of bounded linear operators depending on a continuous real parameter [itex]t[/itex].

 

[WannabeNewton]

You're basically asking whether the function ##A\rightarrow A^{-1}## is continuous where it is defined in ##\mathcal{B}(X,X)## (the normed space of bounded operators on a Banach space). This is in fact true in any Banach algebra. For example, see http://www.iith.ac.in/~rameshg/banachalgebras.pdf Proposition 2.4 at page 10. Now, the function ##t\rightarrow \|A(t)^{-1}\|## is the composition of the continuous functions ##t\rightarrow A(t)##, ##A\rightarrow A^{-1}## and ##A\rightarrow \|A\|##, and is thus continuous.

 

[AxiomOfChoice]

You're basically asking whether the function ##A\rightarrow A^{-1}## is continuous where it is defined in ##\mathcal{B}(X,X)## (the normed space of bounded operators on a Banach space). This is in fact true in any Banach algebra. For example, see http://www.iith.ac.in/~rameshg/banachalgebras.pdf Proposition 2.4 at page 10. Now, the function ##t\rightarrow \|A(t)^{-1}\|## is the composition of the continuous functions ##t\rightarrow A(t)##, ##A\rightarrow A^{-1}## and ##A\rightarrow \|A\|##, and is thus continuous.

This was a very helpful response. Thanks very much!