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I am given this:

Consider a real Banach space $X$ with norm $\|*\|$.

1) Show that the map $x\to \|x\|$ from $X$ to $\mathbb{R}$ is continuous. Is it uniformly continuous?

2) Show that the maps $(x,y)\to x+y$ from $X \times X$ to $X$ and $(c,x)\to cx$ from $\mathbb{R} \times X$ to $X$ are continuous. (On $X \times X$ take the norm $\|(x,y)\|=\|x\|+\|y\|$. On $\mathbb{R} \times X$ take the norm $\|(c,x)\|=|c| + \|x\|$).

I know that normally to show continuity at a point $x_0$, you want to show that $\forall \epsilon > 0, \exists \delta >0$ such that:

$$|x-x_0|\leq \delta \implies |f(x)-f(x_0)|\leq \epsilon$$

So I was thinking that I can set $f(x)=\|x\|$... but other than that, I am confused how to begin attacking this problem. Tips would be appreciated.

Martin Sleziak
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user3784030
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1 Answers1

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Hint: Use the reverse triangle inequality: $$\big|\|x\|-\|y\| \big| \leq \|x-y\|.$$

saz
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  • So using that trick, would it go something like: $\big||x|-|y| \big| \leq |x-y| \leq | \big|x-y\big| | = | \delta | = \delta$ – user3784030 Sep 05 '14 at 06:08
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    @user3784030 Note that $\delta$ is a non-negative real number and not an element of the Banach space $X$, i.e. $|\delta|$ doesn't make sense (the same applies to $| |x-y||$). But I guess you mean the right thing: If $|x-y|<\delta$, then the reverse triangle inequality shows $$| |x|-|y| | \leq |x-y|<\delta.$$ – saz Sep 05 '14 at 06:12
  • So using this, how would I show that $|x-x_0|\leq \delta \implies |f(x)-f(x_0)|\leq \epsilon$ – user3784030 Sep 05 '14 at 06:45
  • @user3784030 Set $\delta=\varepsilon$. – saz Sep 05 '14 at 07:05
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    @user3784030 Alternatively, have a look at the definition of Lipschitz continuity and recall that any Lipschitz continuous function is continuous. – saz Sep 05 '14 at 08:56