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How can I show that $||x|^\alpha-|y|^\alpha|\le|x-y|^\alpha$ for $\alpha\ge0.5$?

I couldn’t even show it for $\alpha=0.5$ How could I go about proving it ?

Parcly Taxel
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user3503589
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  • What have you tried? – Ѕᴀᴀᴅ Oct 09 '18 at 08:23
  • It's not hard to prove $(a+b)^α\leqslant a^α+b^α$ by examining $f(t)=(t+b)^α-t^α$. – Ѕᴀᴀᴅ Oct 09 '18 at 08:24
  • Actually I am not sure how to proceed . I tried to play around a bit but didn’t go far . Even a hint would – user3503589 Oct 09 '18 at 08:24
  • Be highly appreciated . – user3503589 Oct 09 '18 at 08:25
  • False for $\alpha >1$. For $\alpha <1$ don't get carried away by the condition $\alpha \geq 0.5$ The inequality is true for all $\alpha \in (0,1)$. – Kavi Rama Murthy Oct 09 '18 at 08:36
  • Then I am at a loss . I am trying to understand strong uniqueness of the one dimensional SDE $dX_t=\vert X_s\vert^\alpha dW_s$ for $\alpha\ge 0.5 $ as in the paper of Girsanov 1962 via the Yamada Watanabe condition in their paper of 1971 on the dispersion coefficient . Karatzas and Shreve in the Brownian motion book on page 292 in example 2.15 imply that the inequality I asked to prove actually holds . I am going to read more carefully , perhaps I misunderstood their example . Thank you for your comment – user3503589 Oct 09 '18 at 08:49
  • Let $\alpha = 2, y=2, x=1$. Then you have $ \vert 1^2- 2^2 \vert \le \vert 1-2 \vert^2 \quad \rightarrow \quad 3 \le 1 $ which is wrong. Is suggest changing the question with "for $0<\alpha<1$". – Andreas Oct 09 '18 at 08:54
  • Are you sure that in the sources you cited, that you need ALL $\alpha>0.5$ ? – Andreas Oct 09 '18 at 08:56

1 Answers1

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Note: This inequality holds for all $0< p \leq 1$

Define $$d: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}^{\geq 0}$$ by $$d(x,y)=|x-y|^p$$

Now you can show that this function $d$ defines a metric for $0<p \leq 1$. Read the proof here Let $d(x,y)=|x-y|$, when does $|x-y|^p$ define a metric?

Now observe that inequality in the question is nothing but the reverse triangle inequality which says that $$ \forall x, y, z \in \mathbb{R}: \left|{d \left({x, z}\right) - d \left({y, z}\right)}\right| \le d \left({x, y}\right)$$

Put $z=0$ and you are done!!

P.S. for proof of reverse triangle inequality https://proofwiki.org/wiki/Reverse_Triangle_Inequality

Shweta Aggrawal
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