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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?

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2 Answers 2

up vote 1 down vote accepted

Hint: You seem to be almost there. Consider the sequence $\bigl(\,T^{-1}(a_{n })/\Vert T^{-1}(a_{n })\Vert\,\bigr)$, though. This is a sequence in $S$. What can you say about its image under $T$? Note that this sequence has a convergent subsequence...



Solution follows:

Let $x_n=T^{-1}(a_{n })/\Vert T^{-1}(a_{n })\Vert$. Then $(x_n)$ is a sequence of norm one vectors such that $T(x_n)\rightarrow0$ (since $\Vert T^{-1}(a_n) \Vert\rightarrow\infty$). Now, since $S$ is compact, $(x_n)$ has a convergent subsequence $(x_{n_k})$. Moreover, the limit of this subsequence, call it $y$, must have norm one, and in particular $y\ne0$. From the continuity of $T$, we then have $T(y)=0$, contradicting the fact that $T$ is one-to-one.

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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.

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Thank you. I accepted David's answer since it followed the method I had been told, even though yours is more direct. –  rt93 Jun 1 '12 at 0:56

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