# Sums in $\mathbb N^3$

Assume that $a_{n,m,k}\geq 0$ is a sequence, $n,m,k\in \mathbb N$, such that $$\sum_{n,m,k\in \mathbb N} a_{n,m,k}^2 <\infty$$ i.e. it is in $\ell^2 (\mathbb N^3)$.

I want to prove the following:

We can find some path(s) (in the grid of $\mathbb N^3$) escaping to $\infty$, such that the weights $a_{n,m,k}$ of the points $(n,m,k)$ of the path have finite sum. In other words, I'm claiming that we find a path $\gamma$ on the grid $\mathbb N^3$ (with edges added), such that $\gamma$ doesn't stay inside any ball, and such that $$\sum_{a_{n,m,k}\in \gamma} a_{n,m,k} <+\infty$$

For instance, it would suffice to show that

• $\sum_{k} a_{n,m,k}<\infty$ for some fixed $n,m$ or
• $\sum_{m} a_{n,m,k}<\infty$ for some fixed $n,k$ or
• $\sum_{n} a_{n,m,k}<\infty$ for some fixed $m,k$

Do you think that such a statement is true?

This is not possible for $\mathbb N^2$, since the sequence $a_{n,m}= \frac{1}{(n+m)\log(n+m)}$ provides a counterexample, namely it is square summable, but for fixed $n$ or $m$ the sums are infinite.

• You should probably just ask about whether there is a path with finite sum; it's quite easy to find sequences which have no path where two variables remain fixed; e.g., in $d$ dimensions, $$a_{n_1,n_2,\ldots,n_d}=\frac{1}{n_1n_2\ldots n_d}.$$ The question about a finite path, however, is very interesting... (I think it's true) Commented Nov 1, 2014 at 2:19
• If one exhausts $\mathbb{N}^3$ by "shells" of size $\Theta(n^2)$, then the "fake path" connecting the minima $a_l$ on the $l$ th shells will have $(\sum a_l) =\sum(a_l l) (\frac{1}{l}) \leq (\sum (a_l l^2))^\frac{1}{2} (\sum \frac{1}{l^2})^\frac{1}{2} < C \left( \sum a_{n,m,k} \right)^\frac{1}{2} <\infty$. Of course this is not a real "path to infinity". The point being that any potential counterexample would need to have minima "on each shell" positioned far from each other.
– Max
Commented Nov 10, 2014 at 13:28

The question you are asking is related to extremal lengths of graphs. More precisely, Vertex extremal length of a locally finite graph $G$ from a vertex $v$ to $\infty$ (VEL($v \to \infty$)) is defined as $$VEL(v \to \infty) = \sup_m \inf _{\gamma:v \to \infty} \frac{L_m(\gamma)}{Area(m)}$$ where the supremum is taken over all functions $m$ from the vertices of $G$ to $\mathbb R$ such that $Area(m) := \sum_{v \in V(G)}m^2(v) < \infty$, $\gamma$ is a path starting from $v$ going off to $\infty$ and $L_\gamma = \sum_{v \in \gamma} m(v)$. One can similarly define edge extremal lengths (EEL ($v \to \infty$)) by replacing vertices by edges in all the description above. See http://www.iecn.u-nancy.fr/~krikun/pub/fulltext.pdf for more detailed description and references.
Theorem 2.6 in the above link says that on any graph EEL $(v \to \infty) < \infty$ if and only if simple random walk on the graph is recurrent. Also for a bounded degree graph, it is easy to see that EEL$(v \to \infty)<\infty$ is equivalent to saying VEL$(v \to \infty) < \infty$.
It is also easy to see that finiteness of VEL or EEL is independent of the choice of $v$. The question you are asking is exactly if $VEL((0,0,0) \to \infty) < \infty$ in the graph $\mathbb N^3$. It is well known that $\mathbb N^3$ is transient for the simple random walk, hence you can always find a path with finite length. In fact you can find such paths for any $\mathbb N^d$ for the same reason if and only if $d \ge 3$.
Also, you could find the counter example because $\mathbb N^2$ is recurrent.
• Thank you for the references! I was aware of that theorem, and I was actually trying to prove it for $\mathbb N^3$ (or $\mathbb Z^3$), without using the recurrence. Maybe a direct argument does not work then. Commented Nov 11, 2014 at 0:05
• Aah, ok! There might be a direct argument for $\mathbb N^3$, I am not sure. Commented Nov 11, 2014 at 9:14