Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $\langle X_n \rangle$ is a sequence in $\mathbb R^n$ then $\lim\limits_{n\to \infty}X_n=A$. Show that $\lim\limits_{n\to \infty} \| X_n \| = \| A \|$ .

My thought process, $\| X_i \|= (\sum_{i=1}^n x_i^2)^{(1/2)}$ therefore be the multiplicative property of limits $x_i^2 = a_i^2$. Therefore the limit of norm of the components is equivalent to the limit of the corresponding limit of each component of the sequence.

share|improve this question
    
your thought process is correct. The proof relies on the fact that element wise convergence iff convergence and the fact that norm is a continuous function of the components. –  mathemagician Jan 16 '13 at 3:10
add comment

1 Answer

up vote 3 down vote accepted

If $x_n \to a$, then by definition $\|x_n - a\| \to 0$. But we have the subtriangle inequality, so for every $n$ $|\|x_n \| - \|a\| | \le \|x_n - a\|$. The result now follows by the squeeze theorem.

share|improve this answer
    
I've always seen this referred to as the reverse triangle inequality. In any event, +1. –  Benjamin Dickman Jan 16 '13 at 3:08
    
Ahhh,so clever. –  user5208 Jan 16 '13 at 3:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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