Weighted Hamming distance

Basically my question is, what kind of geometry do we get if we use a "weighted" Hamming distance. This is not necessarily Theoretical Computer Science but I think similar things come up sometimes, for instance in randomness extraction.

Define:

$$d(x,y)=$$ the Hamming distance between binary strings $$x$$ and $$y$$ of length $$n$$, $$=$$ the cardinality of $$\{k: x(k)\ne y(k)\}$$.

For a set of strings $$A$$,

$$d(x,A)=\min \{ d(x,y): y\in A\}$$.

The $$r$$-fold boundary of $$A\subseteq \{0,1\}^n$$ is

$$\{x\in\{0,1\}^n: 0 < d(x,A)\le r\}.$$

Balls centered at $$0$$ are given by

$$B(p)=\{x: d(x,0)\le p\},$$ where $$0$$ is the string of $$n$$ many zeroes.

A Hamming-sphere is a set $$H$$ with $$B(p)\subseteq H\subseteq B(p+1)$$. (So it's more like a ball than a sphere, but this is the standard terminology...)

Now, Harper in 1966 showed that for each $$k$$, $$n$$, $$r$$, one can find a Hamming-sphere that has minimal $$r$$-fold boundary among sets of cardinality $$k$$ in $$\\{0,1\\}^n$$. So a ball is a set having minimal boundary -- just like in Euclidean space.

The cardinality of $$B(p)$$ is $${n\choose 0}+\cdots {n\choose p}$$.

The $$r$$-fold boundary of $$B(p)$$ is just the set $$B(p+r)\setminus B(p)$$, which then has cardinality $${n\choose p+1}+\cdots+{n\choose p+r}$$.

So far, so good. But now suppose we replace $$d$$ by a different metric $$D$$: first let $$d_j(x,y)$$ be the Hamming distance between the prefixes of $$x$$ and $$y$$ of length $$j$$, and then $$D(x,y)=\max_{j\le n}\ \frac{d_j(x,y)}{f(j) }$$ where $$0\le f(j)\le j$$. (For example we could have $$f(j)=\sqrt{j}$$, or $$f(j)=j/\log j$$.)

This is supposed to make $$D(x,y)$$ small if the differences of $$x$$ and $$y$$ do not clump together at small values of $$j$$.

Questions

Is the minimum $$r$$-fold boundary (under $$D$$) realized by a $$D$$-ball?

Is there a better definition of $$D$$?

Under the metric $$D$$, what's the minimum size of the $$r$$-fold boundary of a subset of $$\\{0,1\\}^n$$ having cardinality $$k$$? (A reasonable lower bound would be nice.)

(Cross-posted on MathOverflow).