I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.
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22What is that heart stands for? I am not able to understand. – Srijan May 30 '13 at 06:43
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1Wikipedia: Reverse triangle inequality; ProofWiki: Reverse Triangle Inequality. – Martin Sleziak May 30 '13 at 06:45
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2@srijan: I think the elements in LHS is so eager to be less that the RHS ones. Wholeheartedly eager!. – Mikasa May 30 '13 at 06:46
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39Maybe this calls for the love triangle inequality. – Erick Wong May 30 '13 at 06:47
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The same thing for absolute value was asked here: Reverse Triangle Inequality Proof. It is a special case of your question, but the proofs are very similar. – Martin Sleziak May 30 '13 at 06:56
3 Answers
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Observe that
$\lVert x \rVert = \lVert (x -y) +y \rVert \leq \lVert (x -y) \rVert + \lVert y \rVert$
which gives
$\lVert x \rVert - \lVert y \rVert \leq \lVert x -y \rVert$ ... $(1)$
Further,
$-(\lVert x \rVert - \lVert y \rVert ) \leq \lVert (y -x) \rVert = \lVert (x -y) \rVert $... $(2)$
From $(1)$ and $(2)$ result follows.
Srijan
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Use triangle inequality and norm properties to show that $$\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert$$ and $$\lVert y\rVert-\lVert x\rVert\le\lVert x-y\rVert$$
Cameron Buie
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How about applying the triangle inequality to $\parallel x - y + y \parallel$?
Alex Wertheim
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@CameronBuie in cases like these where I am not sure if something is homework, I'd rather give an idea than a solution. Well, a full solution anyway. – Alex Wertheim May 30 '13 at 07:03
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