One-sided derivative of a norm function exists

[Demon117]

Homework Statement

Suppose that || || is a norm on R^n. If p,v are in R^n, show that the one sided derivative

lim( [||p+tv||-||p||]/t, t-->0+) exists.

The Attempt at a Solution

Letting q(t) = ||p+tv||-||p||/t, for s<=t in R, I have already shown that q is bounded below by -||v||. Now I must show that for s<=t, q(s)<=q(t) (i.e. q is increasing). How on Earth is this to be done?

After these two points are shown it follows that lim (q(t), t-->0+) exists and equals inf{ q(t) } (No argument necessary as instructed).

I have done this sort of thing:

q(s)<=q(t) implies (||p+sv||-||p||)t <=(||p+tv||-||p||)s

Then, ||p+sv||t - ||p||t <= ||p+tv||s-||p||s implies ||p+sv||t - ||p+tv||s <= (t-s)||p||. It follows that (t-s)||p|| is positive. So, ||p+sv||t-||p+tv||s >=0.

Then,
0<=||p +sv||t - ||p+tv||s
<=||p||t + ts||v|| - ||p||s - ts||v||
=||p||t - ||p||s

so, ||p||s<=||p||t implies s<=t, which is true by hypothesis. Hence q(s)<=q(t) so q is increasing.

Does that look right?

 

[mighty2000]

It is important to show that q(s) is always less than or equal to q(t) for any s <= t, which you have done by manipulating the inequality and using the fact that s <= t. This shows that q is increasing and therefore has a limit as t approaches 0. Great job!