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Let $M$ be a $3\times2$ matrix. Is it true that for any $x\in\mathbb{R}^{2}$ with $\left\Vert x\right\Vert _{3}=1$ there is some subspace $V$ with dimension $2$ of $\mathbb{R}^{3}$, such that $\left\Vert Mx\right\Vert _{3}\geq\left\Vert M^{T}y\right\Vert _{3/2}$ for all $y\in V$ with $\left\Vert y\right\Vert _{3/2}=1$?

Thanks for any helpful answer.

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Not as you have stated this. –  Charles Matthews Jun 16 '10 at 15:39
    
Charles, sorry that there were some assumptions missing, but they have been added now. –  user6847 Jun 16 '10 at 15:53
    
Dear unknown (yahoo): I have rolled back the recent edits, because the current form seems much more useful to readers. I hope this is not too upsetting. –  S. Carnahan Jun 21 '10 at 17:20
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unknown (yahoo): I hope that you are aware that everyone can view the older versions of the question, so "removing" the question like this doesn't really remove it. A large number of users also have the ability to roll back your removal, which might be a good idea given the effort that the answerers put in. –  Jonas Meyer Nov 25 '10 at 14:55
    
Dear unknown(yahoo): If you have a concern about the continued existence of this question, please contact us at moderators@mathoverflow.net. –  S. Carnahan Nov 25 '10 at 15:22
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4 Answers

No. The inequality you have given implies (by scaling any $y\in V$) that $M^Ty=0$ for all $y\in V$. That is to say, $V\subseteq \operatorname{ker}\left(M^T\right)$. There is no such $V$ if $\operatorname{ker}\left(M^T\right)$ is one-dimensonal.

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Noah, Sorry that the assumptions $\left\Vert x\right\Vert _{3}=1$ and $\left\Vert y\right\Vert _{3/2}=1$ were missing, but they have been added now. –  user6847 Jun 16 '10 at 15:57
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It is true if $M$ is a diagonal matrix. Let $M=\left(\begin{array}{cc} m_{1} & 0\\\ 0 & m_{2}\\\ 0 & 0\end{array}\right)$. Assume that $\left|m_{1}\right|\ge\left|m_{2}\right|$, then for any $x:=\left(\begin{array}{c} x_{1}\\\ x_{2}\end{array}\right)\in\mathbb{R}^{2}$ with $\left\Vert x\right\Vert _{3}=1$, $\left\Vert Mx\right\Vert _{3}=\left(\left|m_{1}x_{1}\right|^{3}+\left|m_{2}x_{2}\right|^{3}\right)^{1/3}\geq\left(\left|m_{2}x_{1}\right|^{3}+\left|m_{2}x_{3}\right|^{3}\right)^{1/3}=\left|m_{2}\right|$. Take $V$ to be the space spanned by $\left(\begin{array}{c} 0\\\ 1\\\ 0\end{array}\right)$ and $\left(\begin{array}{c} 0\\\ 0\\\ 1\end{array}\right)$, then $\left\Vert M^{T}y\right\Vert _{3/2}\leq\left|m_{2}\right|$ for all $y\in V$ with $\left\Vert y\right\Vert _{3/2}=1$. Likewise for the case $\left|m_{1}\right|\le\left|m_{2}\right|$.

But it is still a question for non-diagonal matrices.

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[Edit:]

I think we maybe able to treat this using the duality of the $\ell_3$ and $\ell_{3/2}$ norms.

Observe that it suffices to prove

$$ \inf_{V^2\subset\mathbb{R}^3} \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} \leq \inf_{\|x\|_3 = 1, x\in \mathbb{R}^2} \|Mx\|_3 $$

For a fixed $y$, $$\|M^Ty\|_{3/2} = \sup_{\|x\|_3 = 1} y^TMx$$ For fixed two-space $V$, let $P_V$ denote the orthogonal (rel. $\ell_2$ on $\mathbb{R}^3$) projection to $V$. Then $$ \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} = \sup_{\|y\|_{3/2} = 1} \sup_{\|x\|_3 = 1} y^TP_VMx$$ We can swap the two supremums as the two unit spheres are both compact. So we have $$ \sup_{y\in V, \|y\|_{3/2} = 1} \|M^Ty\|_{3/2} = \sup_{\|x\|_3 = 1} \|P_VMx\|_3$$

Now take $V$ to be the two-space spanned by $v,n$, where $n$ is in the kernel of $M^T$ (which we assume to be only 1 dimensional; if the dimension is bigger, then setting $V$ to be this kernel your inequality is trivial), and $v = Mx_0$ achieves the infimum $$\inf_{\|x\|_3 = 1} \|Mx\|_3$$

Now, the unit ball $B\subset\mathbb{R}^2$ in $\ell_3$ is convex. So is its image under $M$. So the question reduces to whether the supporting hyperplane to $MB$ at $v$ is $\ell_2$ orthogonal to $v$. In the diagonal case this is trivially true. In general I am a bit stuck at the moment.

below is completely wrong. Kept to make the comments to this answer make sense

Let $M = \begin{pmatrix} k&0 \\ k&0 \\ 0 &0\end{pmatrix}$. Then $\inf_{\|x\|_3 = 1}\|Mx\|_3 = 0$. Whereas for any two dimensional subspace $V$, $0 < \sup_{\|y\|_{3/2} = 1, y\in V}\|M^T y\|_{3/2} \propto k$. So the proposed statement is false.

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Willie, in your example, one can choose the plane $V=\left\{ \left(\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\end{array}\right):y_{1}+y_{2}=0\right\} $ , then $\|M^{T}y\|_{3/2}=0$ for all $y\in V$. So the statement still holds. –  user6847 Jun 17 '10 at 1:04
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Yeah, it occurred to me this morning that my example was completely wrong. Something about too much to drink before going to bed can be said... –  Willie Wong Jun 17 '10 at 9:34
    
Willie, that is right, and it is exactly what I have thought about. I think I might be able to work it out now, but thanks a lot. –  user6847 Jun 17 '10 at 23:29
    
Do post an answer if you find one. –  Willie Wong Jun 17 '10 at 23:59
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The inequality can't be true for all M. I agree with Willie Wong's approach and conclusion: the question reduces to whether the tangent to MB at $v$ is orthogonal to $v$. Note that, since $v$ has minimum $l^{3/2}$ norm among all points in the image of the unit $l^3$ ball MB, then the tangent to MB at $v$ is also tangent to the $l^{3/2}$ sphere of radius ||v||3/2, and it is orthogonal to $v$, only in the special case of $v$ in the axes or in the diagonals (in ${\mathbb R}^3$, there are 26 such special points) So, it is sufficient to consider an example of M where $v$ is not one of these special points. In fact, it's quite clear that M can be chosen so that the minimizing point $v$ is any prescribed point, by a continuity argument. Fixing $m_{3,1}=m_{3,2}=0$ makes everything 2 dimensional, and even easier.

[edit] Example with $m_{3,1}=m_{3,2}=0$ (so that one can consider M as a 2x2 matrix). Consider for instance a rotation $M=M_\theta$ of a small angle $\theta$. As above, let $x$ be any point minimizing $\|Mx\|_{3/2}$ on the unit sphere $S:=\{\xi\ : \|\xi\|_3=1\}$. Since $\theta$ is small, $Mx$ has to be close to the corresponding minimizers with $\theta=0$, that is, to one of the 4 unit points in the axes; actually it will be enough to know that $Mx$ is not one of the 4 points on the diagonals. Also, for small $\theta\neq0$, $Mx$ can't be in the axes either (the tangent to $MS$ in a point $v$ which is in the axes is not orthogonal to the axis: therefore $v\neq Mx.$) Since the only points of a $l^{3/2}$ sphere where the tangent is orthogonal to the radius are either on the axes or on the diagonals, one has that the tangent to $MS$ at $Mx$, which by minimality coincides with the tangent to the $l^{3/2}$ sphere at $Mx$, is not orthogonal to $Mx$; this implies that the projection of $MB$ onto the line spanned by $v=Mx$ covers $v$ itself, as already observed in WW's answer.

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Pietro, I have chosen to close the question, but thank you anyway. I think it is true when fixing $m_{3,1}=m_{3,2}=0$ makes everything 2 dimensional, but if you have a counter example, I would be glad to see. –  user6847 Jun 18 '10 at 18:43
    
Well I have added an edit, I hope it's clear. To see the picture, one can exaggerate things, taking the $l^\infty$ and $l^1$ norms instead of the $l^3$ and $l^{3/2}$. Actually I'd say that the inequality is generically never true for any pairs of dual norms, with the only exception of the Euclidean case. –  Pietro Majer Jun 20 '10 at 9:35
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