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show/hide this revision's text 5 the answer is no.

I think the answer to your question is yes. edit: NO

First I'll set some notation. Assume that $X$ is rank $m$. I'll denote by $L$ the m-dimensional plane defined by $X u = y$. Subscripts will denote components of vectors. Instead of $u_{max}$ I'll use $v$. I'll denote by $A_{v}$ the hypercube $0\leq u_i\leq v$ for $1\leq i\leq p$. The problem as stated is about the intersection between $A_v$ and $L$, which is a polytope I'll call $P_v$.

We can rescale the coordinates by taking $u\rightarrow vu'$ so that $A_v$ has side length 1 in the $u'$ coordinates. Under this transformation, $L$ keeps its orientation but is shifted. In particular, $L$ is now defined by $X (vu')=y$, or $X u'=y'$ where $y' = 1/v*y$. As $v$ gets larger and larger, $y'$ gets closer and closer to the origin. Note that if $y$ were the zero vector, your problem is scale invariant and hence has a positive answer.

If $y$ was not the zero vector, then to understand what $P_v$ looks like for large $v$, we need to understand how a slice through the hypercube behaves very close to one of its vertices. Is there a result (for convex polytopes in general?) that tells us that the "shape" of a slice is stable to small translations of the slicing plane when we're close to a vertex? I haven't found any counter-examples in the low-dimensional cases I've (unsystematically) tried.

edit: I spoke way to soon. Consider a plane slicing through the 3-dimensional cube such that the plane makes right angles with the top and bottom faces of the cube. In general the intersection will be a rectangle whose aspect ratio changes and becomes skinnier and skinnier as the plane gets closer to a vertex. So I ask, is there a result that states that the directions of the edges The direction of the intersection will be stable?

If something along these lines holds, then it follows that the largest line longest segment in $P_v$ has a well-defined direction as wellthis rectangle (either of the diagonals) obviously does not stabilize. But is there a positive result lurking here for suitably "generic" planes?
show/hide this revision's text 4 mistake

I think the answer to your question is yes.

First I'll set some notation. Assume that $X$ is rank $m$. I'll denote by $L$ the m-dimensional plane defined by $X u = y$. Subscripts will denote components of vectors. Instead of $u_{max}$ I'll use $v$. I'll denote by $A_{v}$ the hypercube $0\leq u_i\leq v$ for $1\leq i\leq p$. The problem as stated is about the intersection between $A_v$ and $L$, which is a polytope I'll call $P_v$.

We can rescale the coordinates by taking $u\rightarrow vu'$ so that $A_v$ has side length 1 in the $u'$ coordinates. Under this transformation, $L$ keeps its orientation but is shifted. In particular, $L$ is now defined by $X (vu')=y$, or $X u'=y'$ where $y' = 1/v*y$. As $v$ gets larger and larger, $y'$ gets closer and closer to the origin. Note that if $y$ were the zero vector, your problem is scale invariant and hence has a positive answer.

If $y$ was not the zero vector, then to understand what $P_v$ looks like for large $v$, we need to understand how a slice through the hypercube behaves very close to one of its vertices. Is there a result (for convex polytopes in general?) that tells us that the "shape" of a slice is stable to small translations of the slicing plane when we're close to a vertex? I haven't found any counter-examples in the low-dimensional cases I've (unsystematically) tried.

If there is something like this then

edit: I believe spoke way to soon. Consider a plane slicing through the 3-dimensional cube such that the shape plane makes right angles with the top and bottom faces of the cube. In general the intersection $P_v$ for large $v$ will be a rectangle whose aspect ratio changes and becomes scale-invariant from considering low-dimensional examples but I'm not sure how skinnier and skinnier as the plane gets closer to prove ita vertex.

If this So I ask, is truethere a result that states that the directions of the edges of the intersection will be stable?

If something along these lines holds, then it follows that the largest line segment in $P_v$ has a well-defined direction as well.
show/hide this revision's text 3 oops typo

I think the answer to your question is yes.

First I'll set some notation. Assume that $X$ is rank $m$. I'll denote by $L$ the m-dimensional plane defined by $X u = y$. Subscripts will denote components of vectors. Instead of $u_{max}$ I'll use $v$. I'll denote by $A_{v}$ the hypercube $0\leq u_i\leq v$ for $1\leq i\leq p$. The problem as stated is about the intersection between $A_v$ and $L$, which is a polytope I'll call $P_v$.

We can rescale the coordinates by taking $u'\rightarrow vu$ u\rightarrow vu'$ so that $A_v$ has side length 1 in the $u'$ coordinates. Under this transformation, $L$ keeps its orientation but is shifted. In particular, $L$ is now defined by $X (vu')=y$, or $X u'=y'$ where $y' = 1/v*y$. As $v$ gets larger and larger, $y'$ gets closer and closer to the origin. Note that if $y$ were the zero vector, your problem is scale invariant and hence has a positive answer.

If $y$ was not the zero vector, then to understand what $P_v$ looks like for large $v$, we need to understand how a slice through the hypercube behaves very close to one of its vertices. Is there a result (for convex polytopes in general?) that tells us that the "shape" of a slice is stable to small translations of the slicing plane when we're close to a vertex? I haven't found any counter-examples in the low-dimensional cases I've (unsystematically) tried.

If there is something like this then I believe that the shape of the intersection $P_v$ for large $v$ becomes scale-invariant from considering low-dimensional examples but I'm not sure how to prove it.

If this is true, then it follows that the largest line segment in $P_v$ has a well-defined direction as well.
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