Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
9  
What is that heart stands for? I am not able to understand. –  srijan May 30 '13 at 6:43
1  
Wikipedia: Reverse triangle inequality; ProofWiki: Reverse Triangle Inequality. –  Martin Sleziak May 30 '13 at 6:45
2  
@srijan: I think the elements in LHS is so eager to be less that the RHS ones. Wholeheartedly eager!. –  B. S. May 30 '13 at 6:46
19  
Maybe this calls for the love triangle inequality. –  Erick Wong May 30 '13 at 6:47
    
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 6:56
add comment

3 Answers

up vote 4 down vote accepted

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.

share|improve this answer
add comment

Use triangle inequality and norm properties to show that $$\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert$$ and $$\lVert y\rVert-\lVert x\rVert\le\lVert x-y\rVert$$

share|improve this answer
add comment

How about applying the triangle inequality to $\parallel x - y + y \parallel$?

share|improve this answer
    
Fastest way to get the proof done –  Gabriel R. May 30 '13 at 6:55
    
@Gabriel: Half-done, anyway. –  Cameron Buie May 30 '13 at 7:01
    
@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. –  AWertheim May 30 '13 at 7:03
    
$\frac{\text{self improvement}}{\text{homework}}$ –  Trancot May 30 '13 at 7:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.