Triangle Inequality: use to prove convergence (psi function elliptic functions)

Question

1. The problem statement, all variables and given/known data

Attached

I understand the first bound but not the second.

I am fine with the rest of the derivation that follows after these bounds, 2. Relevant equations

I have this as the triangle inequality with a '+' sign enabling me to bound from above:

$|x+y| \leq |x|+|y| $ (1) $|x-y| \geq |x|-|y| $ (2)

and this as the triangle inequality with a '-' sign enabling me to bound from below:

3. The attempt at a solution

So for the first bound we have:

$|z-w| \geq |z| - |w| $

since we have a strict less than inequality for |z| and a strict greater than equality for |w| , both of these are consistent and we indeed loose the equality option in the triangle inequality to get $|z-w| > -R $

I am stuck on the second bound however. 1) I ionly have a upper bound for a subtraction and not a lower via the triangle inequalities. can i get a upper bound from (1) and (2)?

i.e. are you allowewd to do $|z+(-2w)| \leq |z|+|-2w| $?

(even if I am, unlike the lower bound, where it turns out the bound we have on $z$ and $w$ are consistent with the triangle inequality, (enabling us to loose the equality and get strictness) there is contrast between the inequalities in this case. ( both items on the right hand side would need lower bounds (or one upper and one equality) but z has a upper bound).

Many thanks

Answer 1

The triangle inequality is actually a combined inequality. That is, $\;||x|-|y||\le|x+y|\le|x|+|y|.$ In your case, you have written $\;w\;$ for $\;\omega\;$ which is okay. Notice that all that matters is $\;|z|<\frac12|w|\;$ which implies $\;|w|-|z|>\frac12|w|.\;$Now the triangle inequality states $\;||z|-|w||<|z-w|$ and thus, $\;\frac12|w|<|z-w|.\;$ For the rest, $\;|z-2w|=|z+(-2w)|\le|z|+|\!-\!2w|<\frac12|w|+2|w|=\frac52|w|.$