Mathematics Prelims

July 16, 2008

Equivalence of Norms in n dimensions

Filed under: Analysis,Functional Analysis — cjohnson @ 4:08 pm

We say that two norms, \| \cdot \|_1 and \| \cdot \|_2, on the same vector space X are equivalent if there exist \alpha, \beta > 0 such that for every x \in X,

\displaystyle \alpha \| x \|_1 \leq \| x \|_2 \leq \beta \| x \|_1.

In a finite dimensional normed space, all norms are equivalent.

Proof: Suppose X is an n-dimensional space with basis elements \{ e_1, ..., e_n \} and that \| \cdot \|_1 and \| \cdot \|_2 are norms on X.  We know (from an earlier lemma) there exists a c > 0 such that for any x \in X we choose

\displaystyle \| x \|_1 \geq c (| \alpha_1 | + ... + | \alpha_n |)

(Where x = \alpha_1 e_1 + ... + \alpha_n e_n.)

Now consider \| x \|_2 and apply the triangle inequality.

\displaystyle \| x \|_2 \leq k \sum_{i=1}^n | \alpha_i |

Where k = \max \{ \| e_1 \|_2, ..., \| e_n \|_2 \}.  If we then apply our earlier inequality we’re left with

\displaystyle \| x \|_2 \leq \frac{k}{c} \| x \|_1.

If we repeat this process but with \| \cdot \|_1 and \| \cdot \|_2 reversed, we achieve the other inequality.  Since our choice of x \in X was arbitrary, and since our k and c don’t depend on our choice of x, we see that \| \cdot \|_1 and \| \cdot \|_2 are equivalent.  Since these were arbitrary norms, all norms on X are equivalent.



  1. Sadly, there is fatal flaw in this argument here. The quantifiers for x and c have been switched. the statement of the theorem says that there must exist a beta independent of x, however your beta depends on c which depends on x.

    This was a subtle mistake that I fell for at first. I’ve been searching for more elementary proofs of this statement (we have a functional test coming up), and sadly I feel like the essence of this statement requires “work”. So anyone reading this comment, the best I’ve come up with is equating w/ sup norm. Then negating the “hard” direction.

    Comment by Kevin Joyce — November 23, 2008 @ 2:51 pm | Reply

  2. I’m not 100% sure that I understand your concern. I agree that the quantifiers are (or were … fixed now) backwards, but I don’t see what’s wrong aside from that.

    Comment by cjohnson — November 25, 2008 @ 9:24 pm | Reply

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