Lecture 9: Norms are Equivalent in Finite Dimensional Spaces

First we need the following lemma

Theorem

Let $V$ be finite dimensional, then all norms are equivalent.

Proof. Since $V$ is finite dimensional, choose a basis $v_1,\ldots,v_n$. Every vector $v\in V$ can then be written uniquely as

$$ \begin{align*} v=x_1v_1+\cdots+x_nv_n. \end{align*} $$

Define the norm:

$$ \begin{align*} \|v\|_0 = \sum_{i=1}^n|x_i| = |x_1|+\cdots+|x_n|. \end{align*} $$

So, to measure $v$, we add the absolute values of its coordinates in this fixed basis. This is a norm since

Positive definiteness: $\|v\|_0\ge0$, and \[ \|v\|_0=0\iff x_i=0\text{ for every }i\iff v=0. \]
Absolute homogeneity: The coordinates of $\alpha v$ are $\alpha x_i$, so \[ \|\alpha v\|_0=\sum_i|\alpha x_i| =|\alpha|\sum_i|x_i|=|\alpha|\|v\|_0. \]
Triangle inequality: If $w=\sum_i y_i v_i$, then \[ \|v+w\|_0=\sum_i|x_i+y_i| \le\sum_i(|x_i|+|y_i|) =\|v\|_0+\|w\|_0. \]

Thus $\|\cdot\|_0$ is a norm.

Next, take any other norm $\|\cdot\|$ on $V$. We want constants $c,C>0$ such that

$$ \begin{align*} \boxed{ c\|v\|_0\le \|v\|\le C\|v\|_0 } \end{align*} $$

For every $v\in V$. Start with the upper bound. Since $v=\sum_i x_i v_i$ and since $\|\cdot\|$ satisfies the norm axioms, then by the triangle inequality and homogeneity:

$$ \begin{align*} \|v\| =\left\|\sum_i x_i v_i\right\| \le \sum_i |x_i|\,\|v_i\|. \tag{1} \end{align*} $$

This is true for any norm. Now, set

$$ \begin{align*} C=\max\{\|v_1\|,\ldots,\|v_n\|\} = \max_{1\le i\le n}\|v_i\| \end{align*} $$

$C$ is the maximum over the norms of the basis vectors. Hence $\|v_i\|\le C$ for each $i$. Therefore (1) becomes

$$ \begin{align*} \|v\|\le C\sum_i|x_i|=C\|v\|_0. \end{align*} $$

Crucially, $C$ depends only on the fixed basis and the chosen norm, so the same constant works for every $v$. We now have the upper bound:

$$ \begin{align*} \|v\|\le C\|v\|_0. \end{align*} $$

Next we need the lower bound. So we need to show that

$$ \begin{align*} \|v\|\ge c\|v\|_0 \end{align*} $$

Observe that every nonzero vector $v$ can be written as

$$ \begin{align*} v=\|v\|_0\,u \end{align*} $$

where $u=\frac{v}{\|v\|_0}$. Note that $\|u\|_0=1$. By homogeneity of the other norm,

$$ \begin{align*} \|v\|=\|v\|_0\,\|u\|. \end{align*} $$

Thus, it suffices to find a single constant $c>0$ such that $\|u\|\ge c$ for every $u \in V$ with $\|u\|_0=1$. Indeed, for any nonzero $v$, normalization and homogeneity give

$$ \begin{align*} \|v\|=\|v\|_0\left\|\frac{v}{\|v\|_0}\right\| \ge c\|v\|_0. \end{align*} $$

The inequality also holds when $v=0$. We therefore restrict our attention to vectors $u=\sum_{i=1}^n x_i v_i$ satisfying $\|u\|_0=1$, or equivalently,

$$ \begin{align*} \sum_{i=1}^n|x_i|=1. \end{align*} $$

On this set, our desired inequality becomes simply

$$ \begin{align*} \|v\|\ge c>0. \end{align*} $$

References

SFSU Fourier Analysis by Professor Chun-Kit Lai
Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi