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Let $H$ be a Hilbert space. Let $T$ be a linear operator and $R(T)$, $D(T)$, $N(T)$ the range, domain and kernel of $T$, respectively. If $\|Tx\|=\|x\|$ for all $x \in D(T)$, then $T$ is called an $\textit{isometric operator}$.

Let $T$ be an isometric operator. I want to prove that if $\overline{R(T-I)}=H$ (where $I$ is the identity operator) then $N(T-I)=\{0\}$.

My Attempt: Suppose $Tx=x$. I have to prove that $x=0$. The hint is to prove that $\langle (T-I)y, x \rangle=0$ for all $y \in D(T)$, using the fact that $T$ is isometric. I already did this, but I still don't know how to conclude from this that $x=0$. Any help please ? Thanks in advance.

Twnk
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1 Answers1

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You've shown that $\langle (T-I)y,x\rangle=0$ for all $y\in D(T)$. Another way to say that is that $\langle z,x\rangle=0$ for all $z\in R(T-I)$. Now use the fact that $R(T-I)$ is dense in $H$ to conclude that $\langle z,x\rangle=0$ for all $z\in H$, which is what you need to say that $x=0$.

Jordan
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