Proving the completeness of $\mathcal{L}(\mathcal{H})$

Question

Here $\mathcal{L}(\mathcal{H})$ denotes the vector space of all bounded linear operators on a Hilbert space $\mathcal{H}$. We can define a norm on $\mathcal{L}(\mathcal{H})$ by $\|T\| = \inf\{B : \|T(v)\| \leq B \|v\|\}$. Using this, we define a metric $d(T_1, T_2) = \|T_1 - T_2\|$. Under this metric, we wish to prove that $\mathcal{L}(\mathcal{H})$ is complete.

Given a Cauchy sequence $\{T_i\}$ I defined $T(v) = \lim_{i\to\infty} T_i(v)$ (where the limit is obviously taken in $\mathcal{H}$). I have shown that $T \in \mathcal{L}(\mathcal{H})$. How can I show that indeed $T_i \to T$ in $\mathcal{L}(\mathcal{H})$?

It should be noted that this question comes from Stein & Shakarchi's Real Analysis, Chapter 4, Exercise 18. I am not doing this problem for homework. Rather, I'm working the problems in preparation for my upcoming qualifying exam.

Answer 1

Without getting your hands dirty!

Since for A and B being vector spaces normed: $\mathcal{L}(A;B)$ is Banach iff B is Banach, and $\mathcal{H}$ is Banach, then $\mathcal{L(H)}=\mathcal{L(H;H)}$ is Banach.

Answer 2

Let $(T_n)_{n\in\mathbb N}$ be a Cauchy sequence in $\mathcal L(\mathcal H,\mathcal H)$. For each $x\in\mathcal H$, $(T_n(x))_{n\in\mathbb N}$ is Cauchy in $\mathcal H$, since for any $m,n\in\mathbb N$, $$\|T_n(x)-T_m(x)\|\leq\|T_n-T_m\|\|x\|.$$ Given that $\mathcal H$ is complete, $T(x)\equiv \lim_{n\to\infty} T_n(x)$ exists for each $x\in\mathcal H$. For any $x,y\in\mathcal H$ and $a,b\in\mathbb C$, $$T(ax+by)=\lim_{n\to\infty}T_n(ax+by)=a\lim_{n\to\infty} T_n(x)+b\lim_{n\to\infty} T_n(y)=aT(x)+bT(y),$$ so $T$ is a linear function.

Let $N\in\mathbb N$ be so large that $\|T_n-T_N\|<1$ for any $n\geq N$. Fix $x\in\mathcal H$, $x\neq0$, and let $n\in\mathbb N$ be so large that (i) $n\geq N$ and (ii) $\|T_n(x)-T(x)\|<\|x\|$, which is possible due to pointwise convergence and the assumption that $\|x\|>0$. Then, \begin{align*} \|T(x)\|\leq&\,\|T(x)-T_n(x)\|+\|T_n(x)-T_N(x)\|+\|T_N(x)\|\\ \leq&\,\|x\|+\|T_n-T_N\|\|x\|+\|T_N\|\|x\|\leq(2+\|T_N\|)\|x\|. \end{align*} Obviously, this inequality is still true for $x=0$. It follows that $\|T\|$ is bounded. (The crucial part was to make sure that $N$ did not depend on $x$.)

Finally, one can show that $T_n\to T$ in the operator norm. Fix $\varepsilon>0$. Let $N\in\mathbb N$ be so large that $\|T_n-T_m\|<\varepsilon/2$ if $m,n\geq N$. Fix any $x\in\mathcal H$, $x\neq0$ and let $m\in\mathbb N$ so large that (i) $m\geq N$ and (ii) $\|T_m(x)-T(x)\|<\|x\|\varepsilon/2$. Then, if $n\geq N$, the following holds: \begin{align*} \|T_n(x)-T(x)\|\leq&\,\|T_n(x)-T_m(x)\|+\|T_m(x)-T(x)\|\leq\|T_n-T_m\|\|x\|+\frac{\|x\|\varepsilon}{2}\\ \leq&\,\frac{\|x\|\varepsilon}{2}+\frac{\|x\|\varepsilon}{2}=\varepsilon\|x\|. \end{align*} This is true also if $x=0$. It follows that $\|T_n-T\|\leq\varepsilon$ whenever $n\geq N$ (NB: $N$ does not depend on $x$), completing the proof.