📂Banach Space

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

Theorem ¹

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

Explanation

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

Proof

Strategy: If we can show that there exists $c , C >0$ 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. The existence of the maximum and minimum values of $\displaystyle { { \| v \| _{\beta} } \over {\| v \| _{\alpha} } }$ is shown at once through the extreme value theorem.

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If a norm is defined in the finite-dimensional vector space $X$, then there exists a basis $\left\{ e_{1} , \cdots , e_{n} \right\}$. Hence, every vector in $X$ can be represented as $(\lambda_{1} , \cdots , \lambda_{n} ) = \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n}$. Consider the three norms defined in $X$, $\left\| \cdot \right\|_{1}$, $\left\| \cdot \right\|_{2}$, and $\left\| \cdot \right\|$. Especially $\left\| \cdot \right\|$,

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

is defined as the norm by summing up 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.

  Let’s define the function $f : ( \mathbb{C}^{n} , \left\| \cdot \right\| ) \to \mathbb{R}$ as $f( \lambda_{1} , \cdots , \lambda_{n} ) = \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \|_{1}$. And define sequences $\left\{ \lambda_{1}^{(j)} \right\}_{j \in \mathbb{N} } , \cdots , \left\{ \lambda_{n}^{(j)} \right\}_{j \in \mathbb{N} }$ such that each converges to $\lambda_{1} , \cdots , \lambda_{n}$.

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

  That is, when $j \to \infty$, then $\left| f( \lambda_{1}^{(j)} , \cdots , \lambda_{n}^{(j)} ) - f( \lambda_{1} , \cdots , \lambda_{n} ) \right| \to 0$ and thus $f$ is a continuous function.

• Part 2.

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

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

  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 the minimum value $m$ and the maximum value $M$. Therefore,

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

• Part 3.

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

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

  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 the definition of equivalence of norms,

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

• Part 4.

  By showing $\left\| \cdot \right\| \sim \left\| \cdot \right\|_{2}$ using a similar method, by the transitivity of the equivalence relation,

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

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