# Thread: invertible operators

1. ## invertible operators

Let X and Y be normed spaces and let operators $A,B \in L(X,Y)$ continuously invertible (exists $A^{-1}, B^{-1} \in L(X,Y)$). Prove that if
$\mid \mid B-A \mid\mid \le \frac{1}{2\mid\mid A^{-1}\mid\mid},$
then
$\mid\mid B^{-1}-A^{-1}\mid\mid \le 2\mid\mid A^{-1} \mid\mid ^2 \mid\mid B-A\mid\mid.$

2. ## Re: invertible operators

Originally Posted by TheMR
Let X and Y be normed spaces and let operators $A,B \in L(X,Y)$ continuously invertible (exists $A^{-1}, B^{-1} \in L(X,Y)$). Prove that if
$\mid \mid B-A \mid\mid \le \frac{1}{2\mid\mid A^{-1}\mid\mid},$
then $\mid\mid B^{-1}-A^{-1}\mid\mid \le 2\mid\mid A^{-1} \mid\mid ^2 \mid\mid B-A\mid\mid.$
It possible that no one has tried to help is due to notation.
For me $\mathcal{L}(X,Y)$ has always been the set of bounded linear transformations $X\to Y$. Is that the case with this question?

If so is the norm defined as $\left\| A \right\| = {\sup\atop x \ne 0}\dfrac{{\left\| {Ax} \right\|}}{{\left\| x \right\|}}~?$

Please define all notations.

3. ## Re: invertible operators

L(X,Y) is a linear space with the set of continuous linear operators $X \rightarrow Y$.
For $A\in L(X,Y)$:
$\mid \mid A\mid \mid \colon = sup_{x\in B_X} \mid\mid Ax \mid\mid$.

4. ## Re: invertible operators

I don't know how to start. I know that
$\|B-A \| \cdot 2 \| A^{-1}\| \le 1,$
and that
$\| A+B \| \le \| A\| + \| B\|$
and
$\mid \| A\| -\| B\| \mid \ge \| A-B\|$

5. ## Re: invertible operators

At first I should consider the case when X=Y and A=I, the identify operator. In this case $\| B-I \| \le \frac{1}{2}$, so the inverse of B is given by $B^{-1}= \sum_{n=0}^{\infty} (I-B)^n$. I have to show that

$\| B^{-1}-I \| \le 2 \| B-I \| (*)$

but I can't figure it out, can you please help me? I have done the rest with the general case:

\begin{align*}
&\| B-A \| \le \frac{1}{2 \|A^{-1}\| } \mid \cdot \|A^{-1} \| \\
&\| B-A \| \cdot \| A^{-1} \| \le \frac{1}{2} \\
&\| B\cdot A^{-1}-A\cdot A^{-1} \| =\| B\cdot A^{-1} -I \| \le \| B-A\| \cdot \| A^{-1}\|
\le \frac{1}{2}
\end{align*}

Using the previous case (*) with $BA^{-1}$ in place of $B$:
\begin{align*}
&\| (BA^{-1})^{-1} -I\|= \| B^{-1} \cdot (A^{-1})^{-1} -I\| = \| A\cdot B^{-1} -I \|
\le 2 \| BA^{-1}-I \| \\
&\|B^{-1}-A^{-1}\| \le \| A\cdot B^{-1} -I \| \cdot \| A^{-1}\| \le \\
&\le 2 \| BA^{-1}-I \|\cdot \| A^{-1}\|= 2\| A^{-1}\| \cdot \| BA^{-1}-A\cdot A^{-1}\|=\\
&= 2 \| A^{-1}\| \cdot \| A^{-1}(B-A)\| \le 2 \| A^{-1}\|^2 \cdot \| B-A\|
\end{align*}