# Is there a condition for the following consequence?

I have two real vectors $v = (v_1,\ldots,v_n)$ and $u = (u_1,\ldots,u_n)$. I know that the dot product of $u$ and $v$ is larger than $\delta > 0$:

$\langle u,v \rangle \ge \delta$.

What would be an interesting condition on $u$, $v$ or both such that I have $u_i v_i \ge f(\delta)$ for each coordinate $i$ with some real function $f()$?

For example, one condition that I thought about is that for any $j \le n$ we have:

$\sum_{i \neq j} u_i v_i \le \delta/2$

and then we can get that $u_i v_i \ge \delta/2$ for every $i$ using triangle inequality.

You can assume that $||u|| = 1$ and that $||v||$ is bounded by some $M$ (L-2 norms here).

Any help appreciated.

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btw, your example is not true. If for any $j \leq n$ you have $\sum_{i\neq j} u_i v_i \leq \delta / 2$, then summing over $j$ you get $$\sum_j \sum_{i\neq j} u_iv_i = (n-1) \langle u,v\rangle \leq n\delta / 2$$ or equivalently $\langle u,v\rangle \leq n\delta / (2n - 2) < \delta$ if $n > 2$, which contradicts your assumption that the inner product is bounded below by $\delta$. – Willie Wong Feb 14 '11 at 18:06

If I read you correctly, what you are trying to do is not possible in general. But it may be doable if you give more precisely what $u$'s and $v$'s are allowed, as well as what the desired function $f$ is.
Case in point: let $u$ be the unit vector in the $x_1$ direction. Then $u_iv_i = 0$ for any $i\neq 1$. So for any $f(\delta)$ such that $f$ sends positive numbers to strictly positive numbers, the inequality you are looking for is impossible.
In general, it is not generally very doable to try to estimate coordinate dependent quantities (individual $u_iv_i$) by manifestly coordinate independent ones (the dot product). At best you can expect only something trivial in the general case.