📂Banach Space

Proof that All Norms Defined on a Finite Dimensional Vector Space are Equivalent

Theorem¹

All norms defined on a finite-dimensional vector space are equivalent.

Explanation

The fact that all norms defined on Euclidean space are equivalent is a corollary of this theorem.

Proof

Strategy: If we show that there exist $c , C >0$ satisfying $c \| v \| _{\alpha} \le \| v \| _{\beta} \le C \| v \| _{\alpha}$, then the two norms $\left\| \cdot \right\|_{\alpha}$ and $\left\| \cdot \right\|_{\beta}$ are equivalent. Using the extreme value theorem, we show at once that the maximum and minimum of $\displaystyle { { \| v \| _{\beta} } \over {\| v \| _{\alpha} } }$ exist.

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If a norm is defined on the finite-dimensional vector space $X$, then a basis $\left\{ e_{1} , \cdots , e_{n} \right\}$ exists. Therefore, every vector of $X$ can be written as $(\lambda_{1} , \cdots , \lambda_{n} ) = \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n}$. Consider three norms $\left\| \cdot \right\|_{1}$, $\left\| \cdot \right\|_{2}$, and $\left\| \cdot \right\|$ defined on $X$. In particular, $\left\| \cdot \right\|$ is defined by

$$ \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \| : = \sum_{k=1}^{n} | \lambda_{k} | $$

that is, as the norm obtained by summing the absolute values of the coefficients of the linear combination. Now, if we show $\left\| \cdot \right\|_{1} \sim \left\| \cdot \right\|$ and $\left\| \cdot \right\| \sim \left\| \cdot \right\|_{2}$, then by the transitivity of the equivalence relation, any two norms satisfy $\left\| \cdot \right\|_{1} \sim \left\| \cdot \right\|_{2}$.

• Part 1.

  Define the function $f : ( \mathbb{C}^{n} , \left\| \cdot \right\| ) \to \mathbb{R}$ by $f( \lambda_{1} , \cdots , \lambda_{n} ) = \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \|_{1}$. Also, define sequences $\left\{ \lambda_{1}^{(j)} \right\}_{j \in \mathbb{N} } , \cdots , \left\{ \lambda_{n}^{(j)} \right\}_{j \in \mathbb{N} }$ each of which converges to $\lambda_{1} , \cdots , \lambda_{n}$, respectively.

  $$ \begin{align*} & \left| f( \lambda_{1}^{(j)} , \cdots , \lambda_{n}^{(j)} ) - f( \lambda_{1} , \cdots , \lambda_{n} ) \right| \\ =& \left| \| \lambda_{1}^{(j)} e_{1} + \cdots + \lambda_{n}^{(j)} e_{n} \|_{1} - \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \|_{1} \right| \\ \le & \left\| \left( \lambda_{1}^{(j)} e_{1} + \cdots + \lambda_{n}^{(j)} e_{n} \right) - \left( \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \right) \right\|_{1} \\ \le & \sum_{k=1}^{n} \left\| \left( \lambda_{k}^{(j)} e_{k} - \lambda_{k} e_{k} \right) \right\|_{1} \\ \le & \sum_{k=1}^{n} \left| \lambda_{k}^{(j)}- \lambda_{k} \right| \max_{1 \le k \le n} \left\| e_{k} \right\|_{1} \end{align*} $$

  In other words, as $j \to \infty$, we have $\left| f( \lambda_{1}^{(j)} , \cdots , \lambda_{n}^{(j)} ) - f( \lambda_{1} , \cdots , \lambda_{n} ) \right| \to 0$, so $f$ is a continuous function.

• Part 2.

  The set of vectors in $X$ whose sum of the absolute values of the coefficients equals $1$,

  $$ S := \left\{ \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \ : \ \sum_{k=1}^{n} | \lambda_{k}|=1, \lambda_{k} \in \mathbb{C} \right\} $$

  is compact, and since $f$ is continuous, by the extreme value theorem,

  $$ f(S) = \left\{ \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \|_{1} \ : \ \sum_{k=1}^{n} | \lambda_{k}|=1, \lambda_{k} \in \mathbb{C} \right\} $$

  has a minimum $m$ and a maximum $M$. Therefore,

  $$ m \le \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \|_{1} \le M $$

• Part 3.

  For $( \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} ) \in X$ with $\displaystyle \sum_{k=1}^{n} | \alpha_{k}| \ne 1$,

  $$ m \le \left\| {{\alpha_{1} } \over { \sum_{k=1}^{n} | \alpha_{k}|}} e_{1} + \cdots + {{\alpha_{n}} \over { \sum_{k=1}^{n} | \alpha_{k}|}} e_{n} \right\|_{1} \le M $$

  By the properties of norms,

  $$ m \le {{1} \over {\sum_{k=1}^{n} | \alpha_{k}|}} \| \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} \|_{1} \le M $$

  Since $\displaystyle \sum_{k=1}^{n} | \alpha_{k} | = \| \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} \|$,

  $$ m \| \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} \| \le \| \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} \|_{1} \le M \| \alpha_{1} e_{1} + \cdots + \alpha_{n} e_{n} \| $$

  Therefore, according to how the equivalence of norms is defined,

  $$ \left\| \cdot \right\| \sim \left\| \cdot \right\|_{1} $$

• Part 4.

  By showing $\left\| \cdot \right\| \sim \left\| \cdot \right\|_{2}$ in the same manner, by the transitivity of the equivalence relation,

  $$ \left\| \cdot \right\|_{1} \sim \left\| \cdot \right\|_{2} $$

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