# Question about proof that multiplication in Banach algebra is continuous

Here's the proof in my notes:

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.

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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. – h.h.543 Jul 15 '12 at 19:35
I agree with h.h.543. The author even said $\epsilon<1$! – wildildildlife Jul 15 '12 at 21:34

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

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

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As Davide said,

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

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