Let $\mathbb{R}^d, \|\cdot\|_p$ be a finite-dimensional real vector space equipped with the Minkowski $p$-norm $\|\cdot\|_p$. Let $\beta \in \mathbb{R}^d$ be an arbitrary, but fixed, reference point. I am interested in what the boundary between the sets of points closer to (or further from) $\beta$ than the origin looks like when $p \neq 2$.
When $p=2$, one easily finds that the boundary $B_{\beta} =\{x: \|x-\beta\|_2 = \|x\|_2\}$ is the hyperplane through $\beta/2$ defined by $x^{T}\hat{\beta} = \frac{\|\beta\|_{2}}{2}$, where $\hat{\beta}$ is a unit normal vector in the direction of $\beta$.
What does the boundary look like when $p\neq 2$? I'm especially interested in the "corner cases" $p=1$ and $p=\infty$ where geometric intuitions tend to let one down.
Symmetry considerations lead me to suspect that it is still the very same hyperplane when $p \in (3,\infty)$ although - if true - I can't yet figure out how to prove it.
This seems like such a basic question that it surely has been looked at before (perhaps once, because it is easy, or many times because it isn't...).
I'm open to intuitive explanations, but I would especially welcome proofs (or references to proofs) at a level in the neighbourhood of, say, a first course on functional analysis.
Look at $d = 2$. $\|(x,y)\|_1 = |x| + |y|$, while $\|(x,y)\|_\infty = \max \{|x|, |y|\}$. Let $\beta = (\beta_x, \beta_y)$.
To get some intuition, consider what happens for the regular $\|\cdot\|_2$ norm. The set of all points with constant norm is a circle. Draw circles of the same size around both $0$ and $\beta$. When the circles are small enough, they do not overlap. Obviously everything inside either circle is closer to that center than to the other point.
Enlarge the circles, maintaining equal size. Eventually they will touch. That point of tangency is the first point on your hypersurface (a curve), and is the midpoint of the segment connecting the two centers. Increase the shared radius a little more, and the circles now intersect in two points. Those points of intersection are also on your curve. By symmetry, these points are on a line perpendicular to the line segment connecting the centers, but still intersecting it at the midpoint. Thus the curve in question has to be that line.
The "circles" for the $\|\cdot\|_1$ and $\|\cdot\|_\infty$ norms are both squares. For the $\infty$-norm, the sides of the squares are parallel to the axes, while for the $1$-norm, the sides are at $45^\circ$ to the axes. Thus the behavior for the two norms are the same, except that one is rotated $45^\circ$ from the other.
Look at the $\infty$-norm. As with the $2$-norm, we draw equal size squares about $0$ and $\beta$. At first, the squares do not intersect, but if we increase them equally, they eventually will. Where they first touch is on your curve. If $\beta$ lies on one of the two lines through the origin at $45^\circ$ to the axes, these two squares will touch at a single corner, the midpoint between the origin and $\beta$. As the squares enlarge, once again the intersection splits into two points moving off in the perpendicular direction, and you get the very same curve as for the $2$-norm.
But that was only for one very particular situation. When $\beta$ is not on either axis, nor on those $45^\circ$ lines, the initial intersection of the squares is not a single point, but rather a line segment, where the sides of the two squares overlap. Every point on that line segment is equidistant from the two centers, and thus on your curve. When you increase the "radius" even more, the intersection goes back to being two points, where the side of one square intersects the top of the other, and where the side of the second square intersects the bottom of the first. As the "radius" increases, the intersecting edges both move by the same amount, so the intersections move along lines of slope $\pm 1$, at $45^\circ$ to the axes. The final result is a horizontal or vertical line segment, with two rays at $135^\circ$ angles to the line segment, eminating from its two endpoints.
I've saved the weirdest for last. What happens when $\beta$ lies on the $x$ or $y$ axis? As before, the two squares first intersect in a line segment, this time being an entire shared side. But as the size continues to grow, the intersection does not become two points, but rather two line segments, from intersecting edges on opposite sides of the two squares. The endpoints of these segments once again travel along lines of slope $\pm 1$. The result is not a curve, but two V-shaped regions (with right-angles) whose vertices form a square with $0$ and $\beta$, and are connected by a line segment. Every point in the Vs or on the segment is equidistant from $0$ and $\beta$.
For other $p$, the "circle" has curved sides, so I do not believe you get intersections of more than two points at a time. But since $\|\cdot\|_p \to \infty$ as $p \to \infty$, and similarly for $1$, the equidistant surfaces are not going to be hyperplanes.