# can we prove a connection between sum of numbers and $L_2$-like norm?

Let $u$ be a fixed vector of length $K$ and $A$ be a matrix of all positive numbers of size $K \times K$. Let $V$ be a set of vectors.

Let $V(\epsilon) \subseteq V$ be a set of vectors of length $K$ such that:

$0 \le \sup_{v \in V(\epsilon)} u \cdot v < \epsilon$

For an arbitrary vector $w \in V$, define $L(w) = \sum_{i=1,j=1}^K (u_i u_j / A_{ij}) \times w_i w_j$.

Claim: $\sup_{v \in V(\epsilon)} L(v) < C*\epsilon^r$ for some $C,r > 0$.

Proof: $L(v)$ can also be rewritten as: $L(v) = A \colon (u \otimes v) (u \otimes v)^T$

where $\otimes$ is Hadamard product and $\colon$ is the Frobenius inner product. We also know that the Frobenius inner product $A:B = tr(A^t B)$, therefore we have:

$L(v) = tr(A^T (u \otimes v) (u \otimes v)^T)$

which equals (because the reordering of the matrices is a cyclic permutation):

$L(v) = tr((u \otimes v)^T A^T (u \otimes v))$

Here I am stuck. Not sure how to deduce the rest, if it is at all possible. Basically, if I show the following, then the proof is done:

Claim??: let $A$ be a matrix of positive numbers. let $u$ be a vector of all positive numbers. There is a $C$ and an $r$ such that: let $v \in V$ be a vector. let $w = u \otimes v$ be a vector such that $0 < \sum_i w_i < \epsilon$. then $0 \le tr(w A^T w) \le C\epsilon^r$.

However, I don't think the above claim is true. Yet, I have more structure to my problem.

$V$ in fact has the form $V =${$(\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)}) \mid x \in [0,1]$} for an $x^\ast \in [0,1]$

(i.e. $K=2$).

or for $K = 4$ we have

$V =${$(\log \frac{x}{x^\ast}, \log \frac{(1-x)}{(1-x^\ast)},\log \frac{y}{y^\ast}, \log \frac{(1-y)}{(1-y^\ast)}) \mid y,x \in [0,1]$} for an $x^\ast,y^\ast \in [0,1]$

etc. (we can have any even $K$)

could it be true for this $V$?

Thanks for any insight.

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@azimut why such a necromancy? This is 3 years old post. –  tom Nov 12 '13 at 21:29