Let $\mathcal{H}$ be a Hilbert space. $\mathcal{U(H)}$ be the set of all unitary operators on H. Then I want to show that $\mathcal{U(H)}$ is a (norm-) closed subset of the Banach space $\mathcal{L(H)}$, the space of all bounded Linear operators on H.
I have tried taking a sequence of unitary operators on $\mathcal{U(H)}$, but I couldn't conclude. Can somebody please guide me?
Best Answer
Because $L(H)$ is complete, a Cauchy sequence $\{U_n\}$ of unitaries converges to some operator $T$. We also have $$ \|U_n^*-T^*\|=\|(U_n-T)^*\|=\|U_n-T\|\to0. $$ That is $U_n^*\to T^*$. Now \begin{align} \|I-T^*T\| &=\|U_n^*U_n-T^*T\|\\[0.3cm] &\leq \|U_n^*U_n-U_n^*T\|+\|U_n^*T-T^*T\|\\[0.3cm] &\leq\|U_n^*\|\,\|U_n-T\|+\|U_n^*-T^*\|\,\|T\|\\[0.3cm] &=\|U_n-T\|+\|U_n-T\|\,\|T\|\\[0.3cm] &\xrightarrow[n\to\infty]{}0. \end{align} So $T^*T=I$. An entirely similar estimate shows that $TT^*=I$, and hence $T$ is a unitary.