Let $X$ be a Banach space. Let $G$ be the group of invertible linear operators from $X$ to itself. Now my questions are:
- If $G$ is equipped with the operator norm topology, how do you show that it is a topological group?
- If $G$ is equipped with the topology of pointwise convergence, is it still a topological group?
Best Answer
The multiplication map is continuous since it is the restriction of the bounded bilinear map $B(X) \times B(X) \to B(X)$ given by $(S,T) \mapsto ST$. Multiplication is bounded because $\|ST\| \leq \|S\|\|T\|$ by definition of the operator norm.
To check that the inversion map is continuous, it suffices to check that it is continuous in a neighborhood of the identity. But for $S$ with $\|S\| \lt r \lt 1$ we have that $(1-S)^{-1} = \sum_{n=0}^{\infty} S^n$ where the series converges uniformly by majorization with the geometric series, as $\|S^n\| \lt r^n$.
To see this, I'm going to show that neither inversion nor multiplication are strongly continuous for $G$ on an infinite dimensional Banach space $X$.
Adapting item (5) on page 4 in these notes from Hilbert spaces to Banach spaces, we see that the set of operators $\mathscr{N} = \{N \in B(X)\,:\,N^2 = 0\}$ is strongly dense in $B(X)$.
Indeed, a basis for the strong operator topology is given by the sets of the form $$U(S,x_1,\ldots,x_n, \varepsilon) = \{T \in B(X)\,:\,\|(T-S)x_i\| \lt \varepsilon \text{ for all }i\}$$ with $S \in B(X)$, $x_1,\ldots,x_n \in X$ linearly independent and $\varepsilon \gt 0$.
Given such a set $U(S,x_1,\ldots,x_n, \varepsilon)$, pick $y_i$ in such a way that $\|Sx_i - y_i \| \lt \varepsilon$ and that $x_1,\ldots,x_n,y_1,\ldots,y_n$ are linearly independent, so that they span a $2n$-dimensional subspace $Z$. By Hahn-Banach there is a projection $P$ of $X$ with range $Z$ (it can be chosen to be of norm at most $2n$). Now define $Q(x_i) = y_i$, $Q(y_i) = 0$ and set $N = QP$. By definition $N^2 = 0$ and $N \in U(S,x_1,\ldots,x_n,\varepsilon)$.
This establishes that $\mathscr{N}$ is strongly dense in $B(X)$. [Here we proved a bit more than what we actually need, but the extra effort seems minimal.]
For $N$ with $N^2 = 0$ we have $(1+N)(1-N) = 1 = (1-N)(1+N)$ so that $1-N$ is invertible with inverse $1+N$. In particular we have $1\pm N \in G$ for all $N \in \mathscr{N}$.
Now if $N_i \in \mathscr{N}$ is a net converging strongly to an operator $A \in B(X)$ with $\|A\| \lt 1$ and $A^2 \neq 0$ then we have $1\pm N_i \to 1\pm A$ strongly.
Note that $\|A\| \lt 1$ ensures that $1 \pm A \in G$, but $(1-A)^{-1} = \sum_{n=0}^\infty A^n \neq 1+ A$, because $(1+A)(1-A) = 1- A^2 \neq 1$.
Thus, $(1-N_i)^{-1} = 1+N_i$ does not converge to $(1-A)^{-1}$ in the strong operator topology on $B(X)$, so inversion in $G$ is not strongly continuous.
The same construction also shows that multiplication is not strongly continuous, because $(1-N_i) \to 1-A$ and $(1+N_i) \to 1+A$ strongly, while $1 = (1-N_i)(1+N_i)$ does not converge to $(1-A)(1+A) = 1-A^2 \neq 1$.
Finally, in a positive direction, one can show that if the first factor remains bounded, then multiplication is jointly continuous with respect to the strong operator topology.
More precisely, if $(S_i,T_i) \in B(X) \times B(X)$ is a net for which $\sup_i \|S_i\| \lt \infty$ as well as $S_i \to S$ and $T_i \to T$ strongly, then $S_i T_i \to ST$ strongly, because $$ \|(S_i T_i - ST)x\| \leq \|S_i\|\,\|(T_i - T)x\| + \|(S_i - S)Tx\|. $$ This can be used to show that the group of isometric operators $X \to X$ is a topological group in the strong operator topology.