An inequality question 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.
 A: 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.
A: 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.
A: [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. 
A: 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.  
