invertible operators: Let X and Y be normed spaces and let operators $A,B \in L(X,Y)

[theMR]

I could write it that way:

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
$ \| B-A \| \le \frac{1}{2 \|A^{-1}\| },$
then
$\|B^{-1}-A^{-1}\| \le 2 \| A^{-1}\|^2 \| B-A\|.$

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

I don't know how to start.

 

[Deleted member 4993]

View attachment 8029

I could write it that way:

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
$ \| B-A \| \le \frac{1}{2 \|A^{-1}\| },$
then
$\|B^{-1}-A^{-1}\| \le 2 \| A^{-1}\|^2 \| B-A\|.$

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

I don't know how to start.

What are your thoughts?

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[theMR]

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
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*}

 

[Deleted member 4993]

Duplicate post:

http://mathhelpforum.com/advanced-math-topics/274696-invertible-operators.html