# Maximum subset sum of $d$-dimensional vectors

This is a $d$-dimensional generalisation of the post Inequality with Complex Numbers. (See my comment under Robert Israel's answer.)

Generalising Potato's proof for $d$-dimensions, we can show the following theorem:

Theorem. For any positive integer $d$, there exists $c_d > 0$ such that the following holds:

For any $n$ and any $n$-tuple $(v_1, v_2, \ldots, v_n)$ of $d$-dimensional vectors (i.e., $v_i \in \mathbb{R}^d$), there exists a subset $J \subseteq \{ 1, 2, \ldots, n\}$ such that $$\left\| \sum_{j \in J} v_j \right\| \geqslant c_d \cdot \sum_{i=1}^{n} \| v_i \|.$$

I skip my proof of this theorem because it is exactly analogous to Potato's proof; with two modifications: (a) our vectors are in $\mathbb{R}^d$, rather than $\mathbb{C}$ or $\mathbb{R}^2$); (b) I am using the notation $v_i$ in place of $z_i$ here, since it feels more natural in this context.

Now, my question (slightly informally stated) is

Question. For any fixed $d$, what is the largest value $c_d^{\ast}$ of $c_d$ for which the above theorem is true?

If we cannot find the exact value $c_d^{\ast}$, it will be interesting to find nontrivial upper and lower bounds. Working out the details of the above proof, we can show that $c_d$ can be taken to be (at least) $\frac{1}{2d}$; i.e., $c_d^{\ast} \geqslant \frac{1}{2d}$. Can we say anything better?

It is easy to see that $c_1 = \frac{1}{2}$, which matches the above bound for $d=1$. Robert Israel's answer shows that $c_2 = \frac{1}{\pi}$ (exactly!), which is strictly better than $\frac{1}{2 \cdot 2} = \frac{1}{4}$. However, his proof seems to make an essential use of complex numbers that I could not generalise to $d$ dimensions.

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If anyone wants to see the proof I skipped, I will be happy to include it. – Srivatsan Jan 26 '12 at 2:54

## 1 Answer

This simply extends this answer by Robert Israel to $\mathbb{R}^d$.

Define $$R(x)=\left\{\begin{array}{ll}x_1&\text{when }x_1\ge0\\0&\text{when }x_1\lt0\end{array}\right.\tag{1}$$ Integrating $R(x)$ over the sphere of radius $r$ yields $$\omega_{d-2}\;r^d\int_0^{\pi/2}\cos^{d-2}(\theta)\sin(\theta)\;\mathrm{d}\theta=\omega_{d-2}\;\frac{r^d}{d-1}\tag{2}$$ Integrating $1$ over the sphere of radius $r$ yields \begin{align} \omega_{d-2}\;r^{d-1}\int_{-\pi/2}^{\pi/2}\cos^{d-2}(\theta)\;\mathrm{d}\theta &=2\;\omega_{d-2}\;r^{d-1}\int_0^{\pi/2}\cos^{d-2}(\theta)\;\mathrm{d}\theta\\ &=\omega_{d-2}\;r^{d-1}\mathrm{B}\left(\frac{d-1}{2},\frac12\right)\\ &=\omega_{d-2}\;r^{d-1}\frac{\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac d2\right)}\tag{3} \end{align} using the identity $$\int_0^{\pi/2}\sin^{\alpha-1}(t)\;\cos^{\beta-1}(t)\;\mathrm{d}t =\frac12\mathrm{B}\left(\frac{\alpha}{2},\frac{\beta}{2}\right)\tag{4}$$ Therefore, the mean of $R(x)$ over the surface of the sphere of radius $r$ is $(2)$ divided by $(3)$: $$\frac{r\;\Gamma\left(\frac{d}{2}\right)}{(d-1)\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}=\frac{r\;\Gamma\left(\frac{d}{2}\right)}{2\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{d+1}{2}\right)}\tag{5}$$ Thus, the mean of $\displaystyle\sum_{k=1}^nR(x_j)$ over all rotations of the sphere is $$\frac{\Gamma\left(\frac{d}{2}\right)}{2\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{d+1}{2}\right)}\sum_{k=1}^n|x_k|\tag{6}$$ Thus, for some rotation of the sphere, we must have $$\sum_{k=1}^nR(x_k)\ge\frac{\Gamma\left(\frac{d}{2}\right)}{2\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{d+1}{2}\right)}\sum_{k=1}^n|x_k|\tag{7}$$ To see that $(7)$ is the best possible independent of $n$, consider an "even" distribution of $n$ points as $n\to\infty$. This estimates the $(d{-}1)$-dimensional analog of the Riemann sum for the integral in $(2)$.

Therefore, $$c_d=\frac{\Gamma\left(\frac{d}{2}\right)}{2\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{d+1}{2}\right)}\tag{8}$$

As shown in the appendix to this answer, $\displaystyle\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1$. Therefore, asymptotically, $$c_d\sim\dfrac{1}{\sqrt{2\pi d}}\tag{9}$$ which is the geometric mean of $\dfrac{1}{\pi}$ and $\dfrac{1}{2d}$, the upper bound derived in the question.

The first several values of $c_d$ are: $$\begin{array}{c|cccc} d&1&2&3&4&5&6&7&8\\ \hline\\ c_d&\frac{1}{2}&\frac{1}{\pi}&\frac{1}{4}&\frac{2}{3\pi}&\frac{3}{16}&\frac{8}{15\pi}&\frac{15}{96}&\frac{48}{105\pi} \end{array}\tag{10}$$ Furthermore, $(8)$ implies that $$c_{d+2}=\dfrac{d}{d+1}c_d\tag{11}$$

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This is one bloody cool answer! +1. – Jonas Teuwen Jan 26 '12 at 22:16