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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.

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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question closed

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.