5
$\begingroup$

Here's the proof in my notes: enter image description here

Where does the last inequality come from? If I want to show that it's continuous at $((x,y)$ I can use the inverse triangle inequality to get $$ (\|x^\prime\| + \|y\|)\varepsilon \leq (\|x\| + \|y \| + \varepsilon)\varepsilon$$

Thanks.

  • 3
    maybe its a typo, maybe he just felt like writing $1$ rather than $\epsilon$, since it doesn't really make a diference. I don't see how this is important.2012-07-15
  • 1
    I agree with h.h.543. The author even said $\epsilon<1$!2012-07-15

3 Answers 3

4

Since $\varepsilon<1$, $$\color{red}{\lVert x'\rVert}+\lVert y\rVert\leq \color{red}{\lVert x'-x\rVert+\lVert x\rVert} +\lVert y\rVert\leq\color{red}{\varepsilon}+\lVert x\rVert+\lVert y\rVert\leq \color{red}{1}+\lVert x\rVert+\lVert y\rVert.$$

  • 0
    Thanks. May I ask a question which may be a bit off-topic? This proof shows that the multiplication is continuous with respect to the norm, without any reference to the norm condition $||xy||\le||x||||y||$ in the definition of normed algebra. So it means any algebra with a norm has continuous multiplication. Then my question is why do we need the norm condition in the definition of normed algebra ? I thought the norm condition is meant to make multiplication continuous.2017-11-19
2

It comes from the triangle inequality $$||x'||\leq ||x'-x||+||x||,$$ and the hypothesis $$||x'-x||\leq\varepsilon,\quad \varepsilon\in]0,1[$$

0

As Davide said,

$(\|x^\prime\| + \|y\|)\varepsilon \leq (\|x\| + \|y \| + \varepsilon)\varepsilon \leq (\|x\| + \|y \| + 1)\varepsilon$, as $\varepsilon < 1$, hence the given inequality.