# Equivalent Norms

1. Feb 21, 2010

### BSCowboy

1. The problem statement, all variables and given/known data
In $$l_1$$ (each element is a sequence of numbers such that the series converges absolutely) show that no pair of the norms $$||\cdot||_1, \, ||\cdot||_2, \, ||\cdot||_{\infty}$$ are equivalent norms.

2. Relevant equations
$$||x||_1=\sum_{i=1}^{\infty}|x_i|$$
$$||x||_2=\left(\sum_{i=1}^{\infty}|x_i|^2\right)^{\frac{1}{2}}$$
$$||x||_{\infty}=\mbox{ max}\{|x_i|:i=1,2,\ldots\}$$

3. The attempt at a solution
I realize I need to show:
$$\dfrac{||x||_1}{||x||_2}=\infty \text{ or } 0\quad \dfrac{||x||_1}{||x||_{\infty}}=\infty \text{ or } 0\quad \dfrac{||x||_2}{||x||_{\infty}}=\infty \text{ or } 0$$
but I am having problems finding such a sequence and showing this is true.

2. Feb 21, 2010

### Dick

You can't find a SINGLE sequence that will prove they aren't equivalent. You need to find a sequence of sequences. For example, can you find a sequence of sequences x_n where ||x_n||_infinity remains constant as n increases but ||x_n||_1 goes to infinity? That would prove ||_infinity and ||_1 were not equivalent norms, right?

Last edited: Feb 21, 2010
3. Feb 21, 2010

### BSCowboy

How about $$x_n=n^{-1/2}$$

Then: $$||x_n||_1$$ diverges while $$||x_n||_{\infty}=1$$

But, this sequence does not converge absolutely so it wouldn't be in the space, right?

4. Feb 21, 2010

### Dick

That's still a single sequence. You want a sequence of sequences. Define x_n to be the sequence where the first n terms are 1 and the rest are zero. What are ||x_n||_infinity and ||x_n||_1?

5. Feb 21, 2010

### BSCowboy

In that case,
$$||x_n||_1=\infty$$
and
$$||x_n||_{\infty}=1$$

But again, this is not a convergent sequence, so it wouldn't be in the space in the first place right?

Last edited: Feb 21, 2010
6. Feb 21, 2010

### Dick

You aren't quite getting this. x_n isn't a single sequence. It's a different sequence for each value of n. x_1=(1,0,0,0...), x_2=(1,1,0,0,0..), x_3=(1,1,1,0,0...) etc etc. All of those are in l_1. And, yes, ||x_n||_infinity=1. But ||x_n||_1=n, right? What does that tell you about the possibility of the norms being equivalent?

7. Feb 21, 2010

### BSCowboy

You're right, I was being dense (my steady state as of late).

So,

$$\dfrac{||x_n||_1}{||x_n||_{\infty}}=n$$

Then,
$$\dfrac{||x_n||_1}{||x_n||_2}=\dfrac{n}{\sqrt{n}}=\sqrt{n}$$

Lastly,
$$\dfrac{||x_n||_2}{||x_n||_{\infty}}=\sqrt{n}$$

Or, am I still lost?

8. Feb 21, 2010

### Dick

No, I think you are getting it. Since all of those diverge as n->infinity, that means you can't bound any of your norms by a constant times the other norm, right?

9. Feb 21, 2010

### BSCowboy

Right, Thanks Dick. I appreciate it. I guess I kept getting stuck because I wasn't thinking about a sequence of sequences correctly and I kept thinking of $$l_1$$ incorrectly.