Proving Euclidean and supremum norms are equivalent on R²

[gtfitzpatrick]

Homework Statement

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

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?

 

[Hurkyl]

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.

 

[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.

 

[gtfitzpatrick]

thanks for the help lads

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

$$\Rightarrow$$

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

$$\Rightarrow$$

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

a constant?

 

[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?

 

[gtfitzpatrick]

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

 

[Dick]

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

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?

 

[gtfitzpatrick]

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

 

[Dick]

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

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).

 

[gtfitzpatrick]

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

 

[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).

 

[gtfitzpatrick]

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