# Average weighted value of a linear functional over increasing bounded subsets of Z^n

Say you're working within the finite-dimensional free Z-module $\mathbb{Z}^n$, and you want to impose a "norm" on this module. By a "norm" I mean a function $\|·\|: \mathbb{Z}^n \to \mathbb{R}$ which obeys all of the axioms of a norm, but where the domain is the set of elements in the module rather than the set of vectors in a vector space. (I'm not going to keep putting "norm" in scare quotes after this because it's clear what I mean.)

I'm particularly interested in the case where our norm is an $\ell_p$ norm, but also desire to know a general solution to this problem for any arbitrary norm.

Now let $e$ be any element in $\mathbb{Z}^n$, and $f$ be any linear functional in the dual module $\mathbb{Z}^{n*} = Hom(\mathbb{Z}^n, \mathbb{Z})$. Then if we want to find

• $\displaystyle \Theta(f) = \text{max} \left ( \left \lbrace \frac{\left|f(e)\right|}{\|e\|} : \|e\| \neq 0 \right \rbrace \right )$

then it's well-known that $\Theta$ itself has the structure of a norm, and is the dual norm on $\mathbb{Z}^{n*}$. If our norm on $\mathbb{Z}^n$ was an $\ell_p$ norm, this is $\ell_q$ norm dual to it.

Though it may seem odd, we can introduce a limit into the above expression, which leads to the following equivalent expression:

• $\displaystyle \Theta(f) = \lim_{r\to\infty} \text{max} \left ( \left \lbrace \frac{\left|f(e)\right|}{\|e\|} : 0 \lt \|e\| \leq r \right \rbrace \right )$

However, what if rather than the supremum, we want to find the limit of the power mean of these increasingly large (but still finite) sets? That is, say for some real m, we want to find

• $\displaystyle \mu^m(f) = \lim_{r\to\infty} \mu^m \left ( \left \lbrace \frac{\left|f(e)\right|}{\|e\|} : 0 \lt \|e\| \leq r \right \rbrace \right )$

where $\mu^m(S)$ for some finite set $S$ of non-negative reals is the power mean $\displaystyle \left ( \frac{1}{|S|} · \sum_{s \in S} s^m\right )^\frac{1}{m}$?

Given that, my questions are:

• Is there some simple expression for $\mu^m(f)$ in general?
• Will $\mu^m$ in general also have the structure of a norm?
• If not, are there certain special choices of norm on $\mathbb{Z}^n$, and choice of m for $\mu^m$, for which $\mu^m$ does have the structure of a norm?
• If the norm is an $\ell_p$ norm, are there certain special choices of p for which it holds?

Lastly, consider an embedding of $\mathbb{Z}^{n}$ in the obvious way into the vector space $\mathbb{R}^{n}$. This allows us to take $\mu$ and define a new, related function $\hat \mu^m(f) = \left(\frac{1}{\lambda(B_p)}\int_{B_p}\left(\frac{|f(v)|}{\|v\|}\right)^m dv\right)^\frac{1}{m}$, where $v \in \mathbb{R}^n$, $B_p = \left \lbrace v \in \mathbb{R}^n : 0 \lt \|v \| \leq 1 \right \rbrace$, and $\lambda$ is the Lebesgue measure on $\mathbb{R}^n$.

• For any $f \in \mathbb{Z}^{n*}$ (and hence also $\in \mathbb{R}^{n*}$ via the embedding), does $\mu^m(f) = \hat \mu^m(f)$?
• If not, what relationship might exist between $\mu^m(f)$ in the module and $\hat \mu^m(f)$ in the vector space?
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How well does $\left(\frac{1}{\mathrm{vol}(S_p)}\int_{S_p}(v \cdot w)^m dw\right)^\frac{1}{m}$ approximate $\mu^m(f)$? here $S_p = \{w \in \R^n \; | \; \sum_k w_k^p = 1\}$ is the unit $\ell^p$ sphere in $\R^n$ and the functional $f$ is given by $f(e) = v \cdot e$. –  Aaron Hoffman Mar 22 '13 at 21:14
Indeed $\mu^m(f)$ is just the normalized $L^m$ norm of the restriction of $f$ to the unit ball of the $\ell_p$-norm in $\mathbb R^n$ (where $f$ is regarded as a linear function on $\mathbb R^n$). Therefore it is a norm on the dual space. –  Sergei Ivanov Mar 22 '13 at 22:10
Aaron: I'd originally had something in there very similar to that, which I removed to try to keep it more focused. I added it back, or at least my variation of what you wrote (which uses the unit ball instead of the unit sphere and the Lebesgue measure for the volume). Thanks for the suggestion. –  Mike Battaglia Mar 22 '13 at 22:32
Hi Sergei - you caught me mid-edit of my question. Do I understand correctly that you're answering in the affirmative to my second-last bullet point listed, that $\mu^m(f) = \hat \mu^m(f)$ for $f \in \mathbb{Z}^{n*} \in \mathbb{R}^{n*}$? If so, how do you prove this? I see how $\hat \mu^m(f)$ on $\mathbb{R}^n$ could work out nicely like that, but the thing I'm still not sure how to prove is that $\mu^m(f) = \hat \mu^m(f)$ to begin with, so that the property you just mentioned (among other things) holds as well for $\mu^m(f)$ on elements in $\mathbb{Z}^n$. –  Mike Battaglia Mar 22 '13 at 22:55
@Mike: Sorry, I misread the denominator in your formula. In fact, $\mu^m(f)$ is the normalized $L^m$ norm of the function $f(x)/\|x\|$ over the unit ball. In order to get $\hat\mu^m$, one has to replace the denomitator $\|e\|$ by $r$. To see why the sum approaches the integral, rescale the lattice by the $(1/r)$-homothety. Then the sum over the lattice becomes a Riemannian integral sum for the integral over the norm's unit ball. Since the function and the ball are nice, the integral sums converge to the intergal. –  Sergei Ivanov Mar 23 '13 at 11:50