$\lVert a \rVert = 1 = \lVert a^{-1} \rVert \Rightarrow a^* = a^{-1}$

Question

More rigorously, let $A$ be a unital $C^*$-algebra. I'm trying to prove that for $a$ ∈ Inv($A$), $$\lVert a \rVert = 1 = \lVert a^{-1} \rVert \Rightarrow a^* = a^{-1}.$$ To this end, I have already proved the following:

— the absolute value $|a|$ of $a$ constructed using the functional calculus is invertible if $a$ is;

— the element $u := a|a|^{-1}$ is unitary.

This begs the question: in our case, can we always find $b ∈ A$ such that $a = b|b|^{-1}$? If so, I have proved that $a$ is unitary. But perhaps this is not the right way to go about this at all.

Answer 1

By assumption $\|a^*a\|=1$ and $\|a^{-1}(a^{-1})^*\|=1.$ Thus $\|(a^*a)^{-1}\|=1.$ The element $b=a^*a$ is positive, invertible and $\|b\|=\|b^{-1}\|=1.$ Therefore the spectrum of $b$ is contained in $(0,1].$ We have $$\sigma(b^{-1})=\{t^{-1}\,:\, t\in \sigma(b)\}\subset [1,\infty )$$ As $\|b^{-1}\|=1$ the spectrum of $b$ must be equal $\{1\}.$ In that case $b=e$ thus $a^*a=e.$

Answer 2

Using Gelfand-Neimark we may assume that $A$ is a subalgebra of $B(H)$, for some Hilbert space $H$. Then, for every $\xi $ in $H$, we have $$ \|a(\xi )\|\leq \|\xi \| = \|a^{-1}(a(\xi ))\|\leq \|a(\xi )\|. $$ So equality holds throughout and hence $a$ is an isometry, clearly also surjective. Therefore $a$ is unitary.