Let $H$ be a Hilbert space, $T\in L(X)$ with $\|T\|\le 1$ and $A=Id-T^*T$. Suppose that: $\ker(\sqrt{A})=\{x:\|x\|=\|Tx\|\}$ and $\|T\| < 1$. Prove that $A$ is invertible. I try to solve:
$Ax=0 \implies\langle Ax,x\rangle=0 \implies\langle \sqrt{A}\sqrt{A}x,x\rangle=0 \implies\langle \sqrt{A}x,\sqrt{A}x\rangle=0 \implies\|\sqrt{A}x\|=0 \implies \sqrt{A}x=0 \implies x=0$ since $\sqrt{A}$ is injective $\implies$ let $x\in \ker(\sqrt{A}) \implies\|x\|=\|Tx\|$
But $\|Tx\|\le\|T\|$. $\|x\| \implies\|x\|\le\|T\| \|x\|$ then suppose $\|x\|=\|T\| \|x\|$ contradiction since $\|T\|<1$ then $\|x\|<\|T\|$. $\|x\|$ but $\|T\|<1$ hence $\|x\|=0\implies x=0$ hence $\sqrt{A}$ is injective $\implies A$ is injective we still want to prove $A$ is surjective.
Similar to Original Version (Original was even much worse)
${let}\:{H}\:{be}\:{hilbert}\:{space},\:{T}\in{L}\left({H}\right)\:{with}\:\mid\mid{T}\mid\mid\leqslant\mathrm{1} \\ $ ${and}\:{A}={Id}−{T}\ast{T} \\ $
${suppose}\:{that}: \\ $ ${ker}\left(\sqrt{{A}}\right)=\left\{{x}:\mid\mid{x}\mid\mid=\mid\mid{Tx}\mid\mid\right\} \\ $ ${and}\:\mid\mid{T}\mid\mid<\mathrm{1} \\ $
${prove}\:{that}\:{A}\:{is}\:{invertible} \\ $
${I}\:{try}\:{to}\:{solve}: \\ $
${Ax}=\mathrm{0} \\ $
$\Rightarrow\langle{Ax},{x}\rangle=\mathrm{0} \\ $
$\Rightarrow\langle\sqrt{{A}}.\sqrt{{A}}{x},{x}\rangle=\mathrm{0} \\ $
$\Rightarrow\langle\sqrt{{A}}{x},\sqrt{{n}}{x}\rangle=\mathrm{0} \\ $
$\Rightarrow\mid\mid\sqrt{{A}}{x}\mid\mid=\mathrm{0} \\ $
$\Rightarrow\sqrt{{A}}{x}=\mathrm{0} \\ $
$\Rightarrow{x}=\mathrm{0}\:{since}\:\sqrt{{A}}\:{is}\:{injective}\Rightarrow\left\{\right. \\ $
${letx}\in{ker}\left(\sqrt{{A}}\right) \\ $
$\Rightarrow\mid\mid{x}\mid\mid=\mid\mid{Tx}\mid\mid \\ $ ${but}\:\mid\mid{Tx}\mid\mid\leqslant\mid\mid{T}\mid\mid.\mid\mid{x}\mid\mid \\ $
$\Rightarrow\mid\mid{x}\mid\mid\leqslant\mid\mid{T}\mid\mid.\mid\mid{x}\mid\mid \\ $
${then}\:{suppose}\:\mid\mid{x}\mid\mid=\mid\mid{T}\mid\mid.\mid\mid{x}\mid\mid\:\:contradiction\: \\ $
${since}\:\mid\mid{T}\mid\mid<\mathrm{1} \\ $
${then}\:\mid\mid{x}\mid\mid<\mid\mid{T}\mid\mid.\mid\mid{x}\mid\mid \\ $
${but}\:\mid\mid{T}\mid\mid<\mathrm{1}\:{hence}\:\mid\mid{x}\mid\mid=\mathrm{0}\:\Rightarrow{x}=\mathrm{0} \\$
${hence}\:\sqrt{{A}}\:{is}\:{injective}\: \\ $
$\left.\Rightarrow{A}\:{is}\:{injective}\:\right\} \\ $
$ \\ $
${we}\:{still}\:{want}\:{to}\:{prove}\:{A}\:{is}\:{surjective} \\ $