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Consider a $p$-norm acting on vectors in $\mathbb{R}^{2}$ and instead of setting $p\geq 1$, set $0< p < 1$. If we examine the level set at distance 1 away from the origin, $S = \{v : \lVert v \rVert \leq 1, v\in \mathbb{R}^{2}\}$, then this set forms a non-convex set.

From this question: Why is every $p$-norm convex? I know that every norm must be convex and I am interested in building an intuition on how distances between points behave when they are in the "convex part of the interior" versus a "non-convex part of the interior".

If we consider the points $a = (0,1)$ and $b = (1,0)$, then the line segment between $a$ and $b$ is not contained entirely in $S$. What is the shortest path between $a$ and $b$ with respect to the (quasi?)-$p$-norm with $0 < p < 1$?

Based on my intuition, I feel like the shortest path will follow the boundary of $S$ and that if you take a convex subset of $S$, it should obey the Minkowski inequality inside of that set. However on further reflection, I don't think that should necessarily be the case because that would seem to suggest that due to the absolute scaling property, one would be able expand or contract elements in the interior of my level set and make vectors obey or disobey a Minkowski inequality based on that scalar coefficient which doesn't sound right either.

JessicaK
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The shortest path would be to go by horizontal and vertical steps only.

The total length of such a path, from the origin to a point $(x, y)$, would be $|x| + |y|$. If there is any diagonal step, between points that have a difference of $(\Delta_x, \Delta_y)$, that step would have length $(|\Delta_x|^p + |\Delta_y|^p)^{1/p}$ instead of $|\Delta_x| + |\Delta_y|$. If $p < 1$, the diagonal step is longer than the sum of the lengths of a horizontal and a vertical step.

So, the shortest path is not unique, and consists of horizontal and vertical steps. It will not typically follow the boundary of the set.

Robby the Belgian
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