# 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, we
need 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 ${\mathbb{R}}^{n}$ so
we can talk about just those spaces. That is, any finite-dimensional vector
space over $\mathbb{R}$ or $\u2102$ is isomorphic to ${\mathbb{R}}^{n}$
for some $n$ (note that $\u2102$ is just isomorphic to ${\mathbb{R}}^{2}$
as a vector space over $\mathbb{R}$).
To see this, just write any element of the space in of
the basis and then define the isomorphism^{} to take that basis to the standard
basis in ${\mathbb{R}}^{n}$ 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 $\overrightarrow{x}\in {\mathbb{R}}^{n}$ 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 equivalent norms, 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 other
norm is a continuous function^{} with respect to the 2-norm.
Take an arbitrary finite-dimensional space $X$ and
an 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
$\u03f5>0$, 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 of
the above inequality^{}.
Now we note that the triangle inequality^{} immediately also yields the
inequality $|\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 know that 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’t care 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 sphere is 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 shown above, everywhere. ∎

Title | all norms on finite-dimensional vector spaces are equivalent |
---|---|

Canonical name | AllNormsOnFinitedimensionalVectorSpacesAreEquivalent |

Date of creation | 2013-03-22 14:08:54 |

Last modified on | 2013-03-22 14:08:54 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 15 |

Author | jirka (4157) |

Entry type | Theorem |

Classification | msc 46B99 |