all norms on finite-dimensional vector spaces are equivalent

Theorem.

All norms on finite-dimensional vector spaces over $\mathrm{R}$or $\mathrm{C}$are equivalent (http://planetmath.org/EquivalentNorms).

A consequence of this is that there is only one norm induced topology on a finite dimensional vector space. This means that on such a vector space, weneed not worry about what norm we use when we talk about convergence of a sequence of vectors in norm. So a standard use of this theorem is in continuity arguments over finite dimensional vector spaces, and it allows you to pick the most convenient norm for your argument (the Euclidean norm is not always very convenient).

This obviously is not true for infinite dimensional spaces, for example see the different $L^p$ spaces (http://planetmath.org/LpSpace). Note that the reason all this works is because a unit sphere is compact in a finite dimensional vector space, while that is not true in an infinite dimensional one.

Proof.

Any such finite-dimensional space is really just the same as ${ℝ}^n$ sowe can talk about just those spaces. That is, any finite-dimensional vectorspace over $ℝ$ or $ℂ$ is isomorphic to ${ℝ}^n$for some $n$ (note that $ℂ$ is just isomorphic to ${ℝ}^2$as a vector space over $ℝ$).To see this, just write any element of the space in ofthe basis and then define the isomorphism to take that basis to the standardbasis in ${ℝ}^n$ and then extend linearly.

First let’s show that if two norms are equivalent on the unit sphere(all $\overrightarrow{x}$ such that $\parallel \overrightarrow{x}\parallel =1$ with respect to some norm,for example the standard Euclidean norm) then they are equivalent everywhere.We can write any $\overrightarrow{x}\in {ℝ}^n$ as a multiple of some scalar$\gamma \ge 0$ and a vector on the unit sphere, say $\overrightarrow{x_0}$, that is$\overrightarrow{x}=\gamma \overrightarrow{x_0}$.Then when suppose we have two equivalentnorms, say $\parallel \cdot {\parallel }_a$ and $\parallel \cdot {\parallel }_b$, on the unit sphere

$$\alpha {\parallel \overrightarrow{x_0}\parallel }_a\le {\parallel \overrightarrow{x_0}\parallel }_b\le \beta {\parallel \overrightarrow{x_0}\parallel }_a$$

$$\gamma \alpha {\parallel \overrightarrow{x_0}\parallel }_a\le \gamma {\parallel \overrightarrow{x_0}\parallel }_b\le \gamma \beta {\parallel \overrightarrow{x_0}\parallel }_a$$

$$\alpha {\parallel \gamma \overrightarrow{x_0}\parallel }_a\le {\parallel \gamma \overrightarrow{x_0}\parallel }_b\le \beta {\parallel \gamma \overrightarrow{x_0}\parallel }_a$$

$$\alpha {\parallel \overrightarrow{x}\parallel }_a\le {\parallel \overrightarrow{x}\parallel }_b\le \beta {\parallel \overrightarrow{x}\parallel }_a.$$

So the norms are equivalent everywhere.

Suppose we are working with the 2-norm. Now we want to show that any othernorm is a continuous function with respect to the 2-norm.Take an arbitrary finite-dimensional space $X$ andan arbitrary norm $\parallel \cdot \parallel$.Also suppose that ${\{\overrightarrow{b_i}\}}_1^n$ is a basis of $X$ and so an element$\overrightarrow{x}\in X$ may be written as $\overrightarrow{x}={\sum }_1^n{x}_i\overrightarrow{b_i}$.Now given an$ϵ>0$, choose $\delta >0$ such that (the Euclidean distance is less then $\delta$) implies that

In fact we can just choose $\delta$ to be the right side ofthe above inequality.Now we note that the triangle inequality immediately also yields theinequality $|\parallel \overrightarrow{x}\parallel -\parallel \overrightarrow{y}\parallel |\le \parallel \overrightarrow{x}-\overrightarrow{y}\parallel$. So

And so $\parallel \cdot \parallel$ is a continuous function.

Suppose we are given two norms $\parallel \cdot {\parallel }_a$ and $\parallel \cdot {\parallel }_b$, we knowthat they are both continuous functions with respect to the 2-norm.And so the function defined as

$$f(\overrightarrow{x}):=\frac{{\parallel \overrightarrow{x}\parallel }_a}{{\parallel \overrightarrow{x}\parallel }_b}$$

is a continuous function on the unit sphere (with respect to the 2-norm).This function is continuous except perhaps at 0, but we don’tcare about the value at zero. On the unit sphere however $f(\overrightarrow{x})$is continuous and thus achieves a maximum and a minimum since the unit sphereis compact. Let’s call the minimum and maximum,$\alpha$ and $\beta$ respectively.Then for any $\overrightarrow{x}$ on the unit sphere we have

$$\alpha \le f(\overrightarrow{x})\le \beta$$

$$\alpha \le \frac{{\parallel \overrightarrow{x}\parallel }_a}{{\parallel \overrightarrow{x}\parallel }_b}\le \beta$$

$$\alpha {\parallel \overrightarrow{x}\parallel }_b\le {\parallel \overrightarrow{x}\parallel }_a\le \beta {\parallel \overrightarrow{x}\parallel }_b.$$

And so the norms are equivalent on the unit sphere and thus as we shownabove, everywhere.∎