This is a follow-up to this question. Part 3 of that question was stated incorrectly. The correct version is
Show that a continuous invertible linear operator on a normed space has a continuous inverse if the unit sphere is compact
Following a hint from my teacher, my attempt at a solution goes like this:
Let the operator be $T : V \to V$. We know that $T^{-1}$ exists and is linear, and we need to show that it is continuous if the unit sphere is compact. Let $S = \{x \in V : ||x|| = 1\}$ be the unit sphere. Since continuity is equivalent to boundedness for a linear operator, we need to show that $T^{-1}$ is bounded. For a contradiction, suppose $T^{-1}$ is unbounded. Then by definition
$$||T^{-1}|| = \sup_{x \in S}\{||T^{-1}(x)||\} = \infty$$
Let $a_n$ be a sequence in $S$ such that $||T^{-1}(a_n)|| \to \infty$ as $n \to \infty$. Since $S$ is compact, there is a convergent subsequence of this sequence, say $\{a_{n_k}\}$, so $a_{n_k} \to a \in S$.
Now apparently I am supposed to use this subsequence, as well as the continuity of $T$ to arrive at a contradiction, showing that $T^{-1}$ must be bounded. However, I can't work out what to do. In general, I don't really see the relevance of the unit sphere being compact (from a conceptual point of view).
Could anyone help with this?
Best Answer
There's no need to use sequences.
Since $T$ is continuous and $\| \cdot \|$ is continuous, $x \to \|T x\|$ is a continuous real-valued function. Since the unit sphere $S$ is compact, $\|T x\|$ attains a minimum on $S$. The minimum value can't be $0$ because $T$ is one-to-one. So there is $\epsilon > 0$ such that $\|Tx \| \ge \epsilon$ for all $x \in S$. Therefore $\|Tx \| = \|x\| \|T (x/\|x\|)\| \ge \epsilon \|x\|$ for all $x \ne 0$ in your space. In particular, taking $x = T^{-1} y$ this says $\|y\| \ge \epsilon \|T^{-1} y\|$, i.e. $\|T^{-1} y\| \le \epsilon^{-1} \|y\|$, so $T^{-1}$ is bounded.