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Let $v,w\in\mathbb{R}^{2}$ and $v\perp w$. Is it true that $\left\Vert v\right\Vert _{3}\leq\left\Vert v+w\right\Vert _{3}$, in which $\left\Vert \left(x,y\right)\right\Vert _{3}:=\sqrt[3]{\left|x\right|^{3}+\left|y\right|^{3}}$ for $\left(x,y\right)\in\mathbb{R}^{2}$?

Thanks.

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    $\begingroup$ JFYI: your question is probably down-voted because it is not "research level" per se. Though I would've liked to see that people who actually down-voted the question add a comment to this effect when they cast the vote. $\endgroup$
    – Willie Wong
    Commented Jun 14, 2010 at 16:54

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So this asks for the general line L in the plane whether the minimum of the three-norm occurs at the point v obtained by dropping a perpendicular from 0 to L. Not true. Geometrically the tangents to the curves norm = constant rarely have their perpendiculars at the point of tangency passing through 0.

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No. Let $v=(x,y)$ suppose (without loss of generality) that both $x$ and $y$ are $>0$. Then $w=t(y,-x)$ for some $t$ and for small $t$

$$\|v+w\|_3^3= x^3 +y^3 +3t (x^2y-y^2 x) + 3t^2 (xy^2 +yx^2) + t^3 (y^3 -x^3).$$

So long as $x\neq y$ so that $x^2 y - y^2x \neq 0$, the inequality is violated for small $t$.

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  • $\begingroup$ Aren't you assuming $x^2y-y^2x<0$? $\endgroup$
    – Ian Morris
    Commented Jun 14, 2010 at 9:01
  • $\begingroup$ Also you should have $x \neq 0$ and $y \neq 0$. $\endgroup$
    – AgCl
    Commented Jun 14, 2010 at 9:06
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    $\begingroup$ If $x^2y-y^2x>0$, take $t<0$. $\endgroup$
    – Kevin Ventullo
    Commented Jun 14, 2010 at 9:07
  • $\begingroup$ Oh yeah. Whoops. $\endgroup$
    – Ian Morris
    Commented Jun 16, 2010 at 10:57

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