# Continuity of the inverse of a linear operator

1. Jan 5, 2014

### 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 $t$.

Last edited: Jan 5, 2014
2. Jan 9, 2014

### 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.

3. Jan 12, 2014

### AxiomOfChoice

This was a very helpful response. Thanks very much!