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Let $A$ be a unital Banach algebra and $f: Inv(A) \to A$ be the map $a \mapsto a^{-1}$. I'm trying to show that $f$ is differentiable. My idea is to show that the limit of $\delta \to 0$ of $$ {\|(a + \delta a)^{-1} - a^{-1}\| \over |\delta| \|a\|}$$ exists that is, is finite. I have now run out of ideas. Because I got stuck I showed instead that $f$ is continuous which was easy enough. I hoped to use it somehow to show that $f$ is also differentiable but no luck.

Please can someone explain to me how to show that $$ \lim_{\delta \to 0}{\|(a + \delta a)^{-1} - a^{-1}\| \over |\delta| \|a\|}$$ exists?

tom b.
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1 Answers1

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For $a \in \text{Inv}(A)$ let $$ L_a:A\to A,\quad L_a(h)=-a^{-1}ha^{-1}. $$ $L_a$ is obviously a linear continuous map, and for every $h \in A$ with $\|a^{-1}h\|<1$ we have $$ (a+h)^{-1}=(I+a^{-1}h)^{-1}a^{-1}=[I+\sum_{n=1}^\infty(-1)^n(a^{-1}h)^n]a^{-1}=a^{-1}+\sum_{n=1}^\infty(-1)^n(a^{-1}h)^na^{-1}. $$ It follows that \begin{eqnarray} \|f(a+h)-f(a)+a^{-1}ha^{-1}\|&\le&\sum_{n=2}^\infty\|(a^{-1}h)^na^{-1}\|\\ &\le&\|a^{-1}\|\sum_{n=2}^\infty\|a^{-1}h\|^n\\ &=&\|a^{-1}\|\frac{\|a^{-1}h\|^2}{1-\|a^{-1}h\|}. \end{eqnarray} Hence $$ \frac{\|f(a+h)-f(a)-L_a(h)\|}{\|h\|}\le\frac{\|a^{-1}\|^3\|h\|}{1-\|a^{-1}h\|}, $$ i.e. $f$ is differentiable and $$ df_a(h)=-a^{-1}ha^{-1} \quad \forall a \in \text{Inv}(A),\ h \in A. $$

HorizonsMaths
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  • Thanks, how did you come up with $L_a$? – tom b. Jan 01 '14 at 09:09
  • $L_a(h)$ is the linear part in the expansion of $(a+h)^{-1}$. – HorizonsMaths Jan 01 '14 at 20:42
  • Ok, I'm sorry, but why did you figure that the linear part of that expansion is the derivative of $a^{-1}$? – tom b. Jan 02 '14 at 12:31
  • Do you know the definition of the derivative of a map? – HorizonsMaths Jan 02 '14 at 13:37
  • I thought the derivative of a map $f$ at a point $a$ was $$ \lim_{\delta \to 0} {f(a + \delta) - f(a) \over \delta}$$ in the real numbers but then this can be generalized, no? – tom b. Jan 02 '14 at 15:20
  • For a map $f$ between two normed spaces $(X,|\cdot|X)$ and $(Y,|\cdot|_Y)$, the derivative of $f$ at $a \in X$ is a CONTINUOUS LINEAR MAP $L_a$ such that $$\lim{h\to0}\frac{|f(a+h)-f(a)-L_a(h)|_Y}{|h|_X}=0.$$ – HorizonsMaths Jan 02 '14 at 17:17