Continuity of the inverse of a linear operator

[AxiomOfChoice]

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

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

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

 

[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!