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It seems obvious that in an arbitrary normed space $(X, \|\cdot\|)$ a mapping $T$ defined as

$$ T(x) = \begin{cases} \frac{x}{\|x\|} & t \neq 0 \\\\ 0 & t = 0 \end{cases} $$

is continuous everywhere but 0. However, I'm having difficulty formulating a proof. It's quite simple to see that $\|T(x) - T(y)\| \leq 2 \; \forall x,y \neq 0$ and that $\|T(x)\| = 1 \; \forall x \neq 0$, but those properties don't seem to be helpful so far. Where I seem to be having trouble is in finding the right $\delta$ so that $\|x - x_0\| < \delta$ can be brought to bear on $\|T(x) - T(x_0)\|$.

Thanks for any guidance.

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If you consider the function value at some $x+\delta x$ near $x$, you can apply the triangle inequality to the norm in the denominator, and for $\|\delta x\|$ small enough you can bound the resulting fraction to bring $\|\delta x\|$ to the numerator. – joriki Mar 29 '11 at 14:47
up vote 4 down vote accepted

It is probably easiest to do this by combining simple facts you already know:

  1. $X \ni x \mapsto ||x|| \in \mathbb{R}$ is continuous (use the triangle inequality).

  2. $\mathbb{R} \times X \ni (a,x) \mapsto ax \in X$ is continuous.

  3. $\mathbb{R} \setminus \{0\} \ni a \mapsto 1/a \in \mathbb{R}$ is continuous.

Now write $T$ in terms of these.

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So using the fact the the composition of continuous mappings is continuous? Thanks for the insight. – bosmacs Mar 29 '11 at 15:14

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