Bound on a $\chi^2$-type quantity

Question

Suppose $p,q\in([0,1]^{n})^k$ are two vectors such that, for every $1\leq \ell\leq k$, $$ \sum_{j=1}^n p_{\ell,j} = \sum_{j=1}^n q_{\ell,j} = 1 $$ that is, for every fixed $\ell$, $(p_{\ell,j})_{1\leq j\leq n}$ and $(q_{\ell,j})_{1\leq j\leq n}$ are probability distributions. I want to prove the inequality $$ \sum_{j=1}^n\sum_{j'=1}^n \frac{\left(\sum_{\ell=1}^k (p_{\ell,j}-q_{\ell,j})(p_{\ell,j'}-q_{\ell,j'})\right)^2}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})\sum_{\ell=1}^k (p_{\ell,j'}+q_{\ell,j'})} \lesssim n\, \tag{$\dagger$}. $$ Note that by Cauchy-Schwarz, one can get $$\begin{align} \sum_{j=1}^n\sum_{j'=1}^n \frac{\left(\sum_{\ell=1}^k (p_{\ell,j}-q_{\ell,j})(p_{\ell,j'}-q_{\ell,j'})\right)^2}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})\sum_{\ell=1}^k (p_{\ell,j'}+q_{\ell,j'})} &\leq \sum_{j=1}^n\sum_{j'=1}^n \frac{\sum_{\ell=1}^k (p_{\ell,j}-q_{\ell,j})^2\sum_{\ell=1}^k(p_{\ell,j'}-q_{\ell,j'})^2}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})\sum_{\ell=1}^k (p_{\ell,j'}+q_{\ell,j'})}\\ &= \left( \sum_{j=1}^n \frac{\sum_{\ell=1}^k (p_{\ell,j}-q_{\ell,j})^2}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})} \right)^2\\ &\leq \left( \sum_{j=1}^n \max_{\ell}\lvert p_{\ell,j}-q_{\ell,j}\rvert \frac{\sum_{\ell=1}^k \lvert p_{\ell,j}-q_{\ell,j}\rvert}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})} \right)^2\\ & \leq n^2 \end{align}$$ however this is only quadratic in $n$, while my goal is linear. I know how to prove it assuming the distributions are "trivial" (point masses), i.e., when $p,q\in(\{0,1\}^{n})^k$. This was enough for my application, but I feel that $(\dagger)$ should still hold for arbitrary distributions.

Answer 1

You can prove more: if $v_i$ ($i=1,\dots,n$) are vectors in a Hilbert space with pairwise non-negative scalar products and $V=\sum_iv_i$, then $$ \sum_{i,j}\frac{\langle v_i,v_j\rangle^2}{\langle v_i,V\rangle\langle v_j,V\rangle}\le \sum_{i,j}\frac{\langle v_i,v_j\rangle}{\langle v_i,V\rangle}=n\,. $$ To derive your inequality, just consider $v_i=(p_{\ell,i}+q_{\ell,i})_{\ell=1}^k$ and replace $-$ by $+$ in the numerator. You'll get the bound of $4n$ (because $V=(2,2,\dots,2)$ instead of $(1,1,\dots, 1)$ you have in the denominator).

Answer 2

Note that $p_{\ell,j}q_{\ell,j}\geqslant 0$ hence $$\frac{(p_{\ell,j}-q_{\ell,j})^2}{\sum_{\ell'=1}^k (p_{\ell',j}+q_{\ell',j})}\leqslant \frac{ p_{\ell,j}^2+q_{\ell,j}^2}{\sum_{\ell'=1}^k (p_{\ell',j}+q_{\ell',j})} \leqslant p_{\ell,j}+q_{\ell,j} . $$ It follows that $$ \sum_{j=1}^n \frac{\sum_{\ell=1}^k (p_{\ell,j}-q_{\ell,j})^2}{\sum_{\ell=1}^k (p_{\ell,j}+q_{\ell,j})}\leqslant \sum_{j=1}^n\sum_{\ell=1}^k\left(p_{\ell,j}+q_{\ell,j}\right) =2k. $$