-1

Show that if $f_n^2$ converges to $l^2$ then $|f_n|$ tends to $|l|$ as n tends to infinity.

My attempt: Since $f_n^2$ converges to $l^2$, it is bounded. Let $k$ and $K$ be its lower and upper bound respectively. This implies,

$k\leq f_n^2\leq K$

This implies, $\sqrt k\leq f_n\leq \sqrt K$

This implies,

$\sqrt k +l\leq f_n+l\leq \sqrt K+l\leq 2\sqrt K$

Now,

$|f_n^2-l^2|\leq |f_n-l||f_n+l|\leq |f_n-l|(2\sqrt K)<epsilon$

This implies, $|f_n-l|<epsilon/2 \sqrt K$

How do I proceed from here?? Thanks in advance!

Natasha J
  • 825

1 Answers1

2

For all $\epsilon>0$, due $f^2$ converges to $l^2$, exist $\epsilon ' = \epsilon (K+|l|)$ to $|f^2_n-l^2|< \epsilon ' $. This implies $| |f_n|-|l|| = |\sqrt{f^2_n}-\sqrt{l^2}|=\dfrac{|f^{2}_{n}-l^2|}{|f_{n}|+|l|}<\dfrac{\epsilon.(K+|l|)}{|f_{n}|+|l|}<\epsilon; (|f_n|<K)$

Hoang
  • 21
  • 4
  • Thank you so much!! Was my proof wrong? And can you pls explain how you got $|\sqrt(f^2)-\sqrt(l^2)|=\dfrac{|f^2-l^2|}{|f|+|l|}<epsilon$ Thanks – Natasha J Apr 22 '21 at 03:56
  • I guess I asked incorrectly! I meant to ask how did you get $|\sqrt(f^2)-\sqrt(l^2)|=\dfrac{|f^2-l^2|}{|f|+|l|}$? – Natasha J Apr 22 '21 at 14:39