Does the Alternate Triangle Inequality hold if we do not use the outer mod?
That is, instead of $$\Big|\,|x| - |y|\,\Big| \leq |x-y|$$
can we use the following? $$|x| - |y| \leq |x-y|$$
Thanks in advance
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1This is true, but less precise :) – TheSilverDoe Sep 01 '20 at 11:29
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3Yes you can but the latter is a weak version. – Anand Sep 01 '20 at 11:33
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4Note that $|r| \geq r$ for $r \in \mathbb{R}$. – Qi Zhu Sep 01 '20 at 11:42
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Though not so obvious at first glance, the two statements are actually equivalent (and true), so you can use either one you want. – peek-a-boo Sep 01 '20 at 12:06
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1Note that $|a|\leq b$ means $-b \leq a \leq b$ – S.S Sep 01 '20 at 12:20
3 Answers
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$$\begin{align} T_\text{w}&:& |x|-|y| \leq |x-y|\quad&\forall x,y\in S \end{align}$$ is equivalent to $$\begin{align} T_\text{s}&:& \bigl||x|-|y|\bigr| \leq |x-y|\quad&\forall x,y\in S. \end{align}$$
Proof: from $T_\text{s}$ follows $T_\text{w}$ immediately because $a\leq|a|$ holds always. In the other direction, we need to make a case distinction:
- If $|x|\geq|y|$, then $\bigl||x|-|y|\bigr| = |x|-|y|$ and thus $T_\text{s}$ is $T_\text{w}$.
- If $|x|<|y|$, then $$ \bigl||x|-|y|\bigr| = |y|-|x| =: |\tilde{x}|-|\tilde{y}|, $$ which, since $T_\text{w}$ quantifies over all $x,y\in S$, is $$ \leq|\tilde{x}-\tilde{y}| = |y-x| = |x-y|, $$ which was to be shown.
leftaroundabout
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just check that $$ |(x - y) + y| \leqslant |x - y| + |y| \implies |x| - |y| \leqslant |x - y| $$ $\color{red}{\text{The inequality }} \color{blue}{ ||x| - |y|| \leqslant |x - y| \text{ is the general case } }$
Ralph Clausen
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