# Homework Help: Equivalent norms

1. Feb 20, 2010

### gtfitzpatrick

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

Show that the Euclidean and supremum norms are equivalent norms on R2

3. The attempt at a solution

The Euclidean Norm is
$$\left|\left|$$X$$\left|\left|$$1 = $$\sqrt{x1^2 + x2^2}$$

The Supremum norm is
$$\left|\left|$$X$$\left|\left|$$$$\infty$$ = max ($$\left|$$x1$$\left|$$,$$\left|$$x2$$\left|$$)

so for them to be equivalent:

a($$\sqrt{x1^2 + x2^2}$$)$$\leq$$ max ($$\left|$$x1$$\left|$$,$$\left|$$x2$$\left|$$) $$\leq$$ b($$\sqrt{x1^2 + x2^2}$$)

I think i'm going right with this?
im not sure how to work with the supremum norm in the middle?

2. Feb 20, 2010

### Hurkyl

Staff Emeritus
That sounds like a correct algebraic translation of the problem. In more detail, you want to prove:

There exist positive real numbers a and b such that
For any real numbers x1 and x2
$$a \sqrt{x_1^2 + x_2^2} \leq \max \{ |x_1|, |x_2| \} \leq b \sqrt{x_1^2 + x_2^2}$$​

The supremum in the middle is not essential -- with a little effort you can rearrange that inequality so that there are suprema is on the outside and one Euclidean norm in the middle. Really, for the purposes of solving for a and b, you probably just want to look at it as two separate inequalities.

3. Feb 20, 2010

### LCKurtz

Yes, what you have to prove is you can find the a and b. You might start by stating that without loss of generality, you may assume $|x_1| \le |x_2|$, so that:

max{|x1|,|x2|} = |x2|

and work with assumption.

4. Feb 20, 2010

### gtfitzpatrick

$$a \sqrt{x_1^2 + x_2^2} \leq b \sqrt{x_1^2 + x_2^2} \geq \max \{ |x_1|, |x_2| \} }$$

$$\Rightarrow$$

$$a \sqrt{x_1^2 + x_2^2} \leq b \sqrt{x_1^2 + x_2^2} \geq |x_2| }$$

$$\Rightarrow$$

$$a \leq b \geq |x_2|/\sqrt{x_1^2 + x_2^2} }$$

a constant?

5. Feb 20, 2010

### Dick

Just draw a right triangle. Make the lengths of the legs x1 and x2. sqrt(x1^2+x2^2)=h is the length of the hypotenuse. In terms of h, what's the longest the longest leg can be? What's the shortest the longest leg can be?

6. Feb 20, 2010

### gtfitzpatrick

the shortest the longest leg can be is x1=x2 so then x2 = $$\sqrt{(h^2)/2}$$ ?

7. Feb 20, 2010

### Dick

Exactly. h/sqrt(2). How about the longest the longest leg can be? So the max norm is bounded by constants times the euclidean norm, right?

8. Feb 20, 2010

### gtfitzpatrick

the longest the longest leg can be is x1 = 0 so then x2 = h?

9. Feb 20, 2010

### Dick

I would have said 'the longest the longest can be is h'. But I think you've got the idea. If h is the hypotenuse, the longest leg is between h and h/sqrt(2).

10. Feb 20, 2010

### gtfitzpatrick

ok, so the longest leg is between h and h/sqrt(2), i get that with the triangle but im not sure how it shows the norms are equivalent?sorry and thanks for the help!

11. Feb 20, 2010

### Dick

The hypotenuse is the euclidean norm of the point (x1,x2) in R^2. The longest leg is the max norm of (x1,x2).

12. Feb 20, 2010

### gtfitzpatrick

oh god,yes-thanks a million...that actually really,really explains it to me. Damn thanks a million.Thanks dick. Im actually happy now