# Norm ineqaulity

## Homework Statement

Show that

$$\frac{\Vert X(u+v) \Vert}{\Vert u+v \Vert} \leq \max \{ \frac{\Vert Xu \Vert}{\Vert u \Vert}, \frac{\Vert Xv \Vert}{\Vert v \Vert} \}$$

## The Attempt at a Solution

Tried to rewrite the max statement as an inequality (without loss of genreality). Then However I can't really get anyway with it since
when I try to estimate the numerator or the denominator independently (triangle inequality, ...) I get a bound which is too high and I don't really know how to estimate both simultaniuously.

thx

this doesn't seem to be working. counter example:
X=[[2,0],[0,1]] (x-coordinate is doubled)
u=(1,1)
v=(1,-1)
u+v=(2,0)
rescaled to the unit circle:
u/|u|=(1/sqrt(2),1/sqrt(2))
v/|v|=(1/sqrt(2),-1/sqrt(2))
u+v/|u+v|=(1,0)
applying the matrix X to these:
X(u/|u|)=(2/sqrt(2),1/sqrt(2))
X(v/|u|)=(2/sqrt(2),-1/sqrt(2))
X(u+v/|u+v|)=(2,0)
but the lengths of the first two are both sqrt(2.5) < 2

Thanks for your answer. Now I'm slightly confused. Actually the example is taken from "Introduction to Applied Nonlinear Dynamical Systems". where it is stated that

For any vectors $$f,g\in\mathbb{R}^n$$
$$\chi(f+g) \leq \max\{\chi(f),\chi(g)\}$$

where $$\chi$$ is the Lyapunov exponent given by.

$$\chi(X,e) = \lim_{t\to\infty} \frac{1}{t} \log \frac{\vert Xe\vert}{\vert e \vert}$$

where X in general does depend on t.

Since the logarithm is a montonuous function and i have to show the behavior for all $$t$$ such that it holds in the limit (or at least for some t>T). The book states that this follows readily from the defintion....

thx