Question about Normed Linear Spaces

  • Thread starter Thread starter Oster
  • Start date Start date
  • Tags Tags
    Linear
Oster
Messages
84
Reaction score
0
Statement: V is a finite dimensional vector space with basis {ei} (i goes from 1 to n). V has a norm || || defined on it(not necessarily induced by an inner product). Let x=Ʃxiei belong to V. I want to show that ||x|| ≥ ||xiei|| for any fixed i.

I'm not entirely sure this result is correct. But i remember seeing something similar in a text a while ago.
I know all the properties of a norm but I'm not sure how to proceed. I don't know how the independence of the basis vectors will fit into the proof.
 
Physics news on Phys.org
It isn't true. In ##R^2## consider the basis ##e_1=<1,0>,\ e_2=<-1,.1>##. Then let ##x=1e_1+1e_2=<0,.1>##. Then ##\|x\|=.1<1\|e_1\|##.
 
Thank you!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top