Derivative of Inverse Function on Banach Space

Question

During the semester, I ran into a problem that dealt with finding the derivative of the inverse function on a Banach space (where $V$, the vector space is a Banach algebra and the field $F = R$ or $C$).

I had managed to show that the set of invertible elements of the Banach algebra was open, and then I needed to find the derivative of $f(x) = x^{-1}$ where $f : GL(V) \to GL(V)$. Here is what I have thought so far:

So the derivative satisfies $\lim_{h \to 0} |\frac{(x+h)^{-1} - x^{-1} - \lambda(h)}{|h|}|$. We have that $(x+h)^{-1} - x^{-1} = x^{-1}((e + x^{-1}h)^{-1} - 1) = x^{-1}(\sum_{j=0}^{\infty} (x^{-1}h)^{j} - 1) = x^{-1}(\sum_{j=1}^{\infty} (x^{-1}h)^{j})$. This is as far as I have been able to get. Any hints or helps would be greatly appreciated.

Working off the hint from Thomas Shelby: We show that $f$ is continuous on $GL(V)$. Given some $x \in GL(V)$ (let $B(x, h) \subseteq GL(V)$ and an $\epsilon > 0$ we have two cases. If $||x||\epsilon \geq 1$ then choose $\delta = \min(1, h)$ else choose $\delta = \min(\frac{||x||^2\epsilon}{2(1-||x||\epsilon)}, h)$. Then we have for $y$ such that $||x - y|| < \delta$ that $|f(x) - f(y)|| = ||x^{-1}(y-x)y^{-1}|| = ||x^{-1}||*||y^{-1}||*||x-y|| = \frac{||x-y||}{||x||*||y||}$ and since $||y|| \leq ||x - y|| + ||x||$ by triangle inequality we have that this is less than or equal to $\frac{\delta}{||x||(||x|| + \delta)} < \frac{\frac{||x||^2\epsilon}{1-||x||\epsilon}}{\frac{||x||^2}{1-||x||\epsilon}} = \epsilon$.

To conclude then we note that as Thomas Shelby wrote, $||\frac{(x+h)^{-1} - x^{-1} + x^{-1}hx}{||h||}|| = \frac{||(x+h)^{-1} - x^{-1}||*||h||*||x^{-1}||}{||h||} = ||(x+h)^{-1} - x^{-1}||*||x^{-1}||$ which goes to $0$ as $h \to 0$ by continuity of $f$.

Answer 1

Hint:

Choose $h$ small enough so that $x+h $ is invertible. Then \begin{align*} (x+h)^{-1}-x^{-1}&=(x+h)^{-1}[x-(x+h)]x^{-1}\\ &=-(x+h)^{-1}hx^{-1}. \end{align*} So, \begin{align*} (x+h)^{-1}-x^{-1}+x^{-1}hx^{-1}&=-(x+h)^{-1}hx^{-1}+x^{-1}hx^{-1}\\ &=[x^{-1}-(x+h)^{-1}]hx^{-1}. \end{align*}

Now, show that $Df(x)h=-x^{-1}hx^{-1}$ using the continuity of $f$.

Answer 2

For continuity of the inversion, you can use a trick together with the geometric series.

Note that $(e- x^{-1}(x-y))^{-1} = \sum_{k=0}^\infty (x^{-1} (x-y))^k$ as soon as $|y-x|$ is small enough.

And since $y=x (e - x^{-1} (x-y))$, we have then

$|y^{-1}-x^{-1}|=|\sum_{k=0}^\infty (x^{-1})^{k+1} (x-y)^k-x^{-1}|=$

$|\sum_{k=1}^\infty (x^{-1})^{k+1} (x-y)^k|\le |x^{-1}|\sum_{k=1}^\infty |(x^{-1})^k (x-y)^k|=$

$|x^{-1}|\left(\frac{1}{1-|(x^{-1})(x-y)|}-1\right)\to 0$ as $y\to x$.