1
$\begingroup$

Let $x=(x_1,...x_m) \in R^m$, prove that $$ \frac{\sqrt{m}}{m} \sum_{i=1}^m |x_i| \leq |x| \leq \sum_{i=1}^m |x_i|$$, where $|x|=\sum_{i=1}^m x_i$

The proof goes by the following:

Since $$(\sum_{i=1}^m |x_i|)^2=\sum_{i=1}^m x_i^2+2 \cdot \sum_{i=1_{m\geq j >i}}^m |x_i| |x_j| \leq |x|^2 \tag{1}$$

We get $$|x|=(\sum_{i=1}^m |x_i|^2)^{\frac{1}{2}} \leq \sum_{i=1}^m |x_i| \tag{2}$$

By Cauchy inequality, $$\sum_{i=1}^m |x_i|=\sum_{i=1}^m 1 \cdot |x_i| \leq \sum_{i=1}^m 1^2 \cdot |x_i|^2)^\frac{1}{2} \leq \sqrt{m}(\sum_{i=1}^m (x_i)^2)^\frac{1}{2}=\sqrt{m} |x|$$, hence proved.

Can someone explain the tags 1 and 2 please?

$\endgroup$
1
$\begingroup$

In the first it means the following. $$|x|=\left|\sum_{i=1}^mx_i\right|=\sqrt{\left(\sum_{i=1}^mx_i\right)^2}=\sqrt{\sum_{i=1}^mx_i^2+2\sum_{1\leq i<j\leq m}x_ix_j}\leq$$ $$\leq\sqrt{\sum_{i=1}^mx_i^2+2\sum_{1\leq i<j\leq m}|x_ix_j|}=\sqrt{\sum_{i=1}^m|x_i|^2+2\sum_{1\leq i<j\leq m}|x_i||x_j|}=\sum_{i=1}^m|x_i|.$$ I think it's better to see it by the triangle inequality.

The left inequality is wrong. Try $x=0$ and $x_1=1$.

$\endgroup$

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.