let $X$ be a complex Hilbert space with the inner product $\langle\cdot, \cdot \rangle$ and norm $||\cdot||$. Suppose $T$ is a bounded linear operator on $X$. Prove if $X_0$ is a dense linear subspace of $X$ that $||T(a)||=||a||$ for every element $a \in X_0$ and range $R(T)$ of $T$ is dense in $X$, then $T$ is unitary.
For me the dense set definition is every open set at least contain a point of dense set, I don't know what is relationship between dense set and unitary. Hope to get proof of this question, thanks.
A unitary operator is simply an isometry which is surjective. Note that $T$ is a bounded operator, so the equation $\|Tx\|=\|x\|$ for $x \in X_0$ extends to $X$. To show that $T$ is unitary it is enough to show that the range is closed (because a closed set which also dense is equal to the whole space). Let $Tx_n \to y$. Then $\|x_n-x_m\| =\|Tx_n-Tx_m\|\to 0$. By completeness of $X$ there exists $x \in X$ such that $x_n \to x$. But then $y=\lim Tx_n =Tx$ since $T$ is continuous. This completes the proof.