3
$\begingroup$

If $\lvert a_i\rvert < 1$, $\lambda_i\geq 0$ for $i = 1,\ldots,n$ and $\lambda_1 + \lambda_2 + \cdots + \lambda_n = 1$, show that $$ \lvert\lambda_1a_1 + \lambda_2a_2 + \cdots + \lambda_na_n\rvert < 1. $$


Since $\sum_{i = 1}^n\lambda_i = 1$ and $\lambda_i\geq 0$, we have that $0\leq \lambda_i < 1$. By Cauchy's inequality, $$ \lvert\lambda_1a_1 + \lambda_2a_2 + \cdots + \lambda_na_n\rvert^2 \leq\bigl(\lvert a_1\rvert^2 + \cdots + \lvert a_n\rvert^2\bigr) \bigl(\lvert\lambda_1\rvert^2 + \cdots + \lvert\lambda_n\rvert^2\bigr) $$ Since $\lambda_i^2 < \lambda_i < 1$, we have $\lvert\lambda_1\rvert^2 + \cdots + \lvert\lambda_n\rvert^2 < 1$ and $$ < \lvert a_1\rvert^2 + \cdots + \lvert a_n\rvert^2\tag{1} $$ How do I show equation $(1)$ is less than $1$?

$\endgroup$
6
$\begingroup$

By the triangle inequality we have $$|\lambda_1 a_1 + ... +\lambda_n a_n|\leq |\lambda_1||a_1|+...+|\lambda_n||a_n|$$

Since each $|a_i|<1$ and each $\lambda_i$ is non-negative, we have $|\lambda_i|= \lambda_i$. Then $$|\lambda_1||a_1|+...+|\lambda_n||a_n| < \sum_i \lambda_i =1$$

Since you tagged this as complex analysis, I assume you meant for the $a_i$ be complex numbers and the $\lambda_i$ to be non-negative real numbers.

$\endgroup$
  • $\begingroup$ Yes the $a_i\in\mathbb{C}$ and $\lambda_i$ weren't stated but I assumed they were real. $\endgroup$ – dustin Jan 25 '15 at 2:45
  • 3
    $\begingroup$ I suppose the $\lambda_i$ are implicitly real, otherwise the statement $\lambda_i\geq 0$ wouldn't make sense. $\endgroup$ – Sergio Da Silva Jan 25 '15 at 2:47
  • $\begingroup$ That was my thought as well. $\endgroup$ – dustin Jan 25 '15 at 2:48
4
$\begingroup$

I would like to suggest "different" approach. First off as noted in one of the comments $\lambda_i$ must be real since $\mathbb{C}$ is not an ordered field. On the another hand $\mathbb{C}$ is a vector space over the field of real numbers. So I am going to translate your original statement into a geometric interpretation. Imagine that you pick $n$ points in the complex plain each of which is inside of the unit disk with the centre at the origin. $\lambda_1a_1+...+\lambda_na_n$ for the given condition $\lambda_i\geq 0$ and $\lambda_1+...+\lambda_n=1$ is the convex combination of those points, a simplex around the origin if you like having vertices inside the unit disk. Pick up any point from that simplex. Is it possible that the point lie outside of the unit disk?

$\endgroup$
  • $\begingroup$ This is a good point. I was going to mention viewing the problem from the simplicial perspective, but the way you have phrased it is better than I would have explained. $\endgroup$ – Sergio Da Silva Jan 25 '15 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.